TITLE: Truly intuitive geometric interpretation for the transpose of a square matrix
QUESTION [12 upvotes]: I'm looking for an easily understandable interpretation for a transpose of a square matrix A. An intuitive visual demonstration, how $A^{T}$ relates to A. I want to be able to instantly visualize in my mind what I'm doing to the space when transposing the vectors of a matrix.
From experience, understanding linear algebra concepts in two dimensions is often enough to understand concepts in any higher dimension, so an explanation for two dimensional spaces should be enough I think.
All explanations I found so far were not intuitive enough, as I want to be able to instantly imagine (and draw) how $A^{T}$ looks like given A. I'm not a mathematician btw.
Here is what I found so far (but not intuitive enough for me)

(Ax)$\cdot$y=$(Ax)^{T}$y=$x^{T}A^{T}$y=x$\cdot$$A^{T}$y

As far I understand dot product is a projection (x onto y, y onto x, both interpretations have the same result) followed by a scaling by the length of the other vector.
This would mean that mapping x into space A and projecting y onto the result is the same as mapping y into the space of $A^{T}$, then projecting the unmapped x into $A^{T}$y
So $A^{T}$ is the specific space B for any pair of vectors (x,y) such that Ax$\cdot$y=x$\cdot$By
This doesn't tell me instantly how $A^{T}$ drawn as vectors would look like based on A drawn as vectors.

"reassigning dimensions"

This one is hard to explain so let me do this with a drawing:
parallel projections
This explanation is much more visual, but far too messy to do it in my head instantly. There are also multiple ways I could have rotated and arranged the vectors around the result $A^{T}$ which is represented in the middle. Also, it doesn't feel like it makes me truly understand the transposing of matrices, especially in higher dimensions.

some kind of weird rotation

Symmetrical matrices can be decomposed into a rotation, scaling along eigenvectors $\Lambda$ and a rotation back
A=R$\Lambda$$R^{T}$
So in this specific case, the transpose is a rotation in the opposite direction of the original. I don't know how to generalize that into arbitrary matrices. I'm wildly guessing that if A is not symmetric any more, $R^{T}$ must also include some additional operations besides rotation.
Can anyone help me to find a way to easily and instantly imagine/draw how $A^{T}$ looks like given A in two dimensional space? (In a way of understanding that is generalizable into higher dimensions)
Edit 1:
While working on the problem I was curious to see what B in

$BA=A^{T}$

looks like. B would describe what needs to be done to A in order to geometrically transpose it. My temporary result looks interesting but I'm still trying to bring it to an interpretable form. If we assume the following indexing order
$$A=
        \begin{bmatrix}
        a_{11} & a_{12} \\
        a_{21} & a_{22} \\
        \end{bmatrix}
$$
and $det(A)\neq0$ then
$$B=\frac{1}{det(A)}
        \begin{bmatrix}
        a_{11} a_{22} - a_{21}^2 & a_{11} (a_{21} - a_{12})  \\
        a_{22} (a_{12} - a_{21}) & a_{11} a_{22} - a_{12}^2 \\
        \end{bmatrix}
$$
What's visible on the first sight is that $\frac{1}{det(A)}$ causes scaling such that the area becomes exactly 1 (before applying the actual matrix).
B must also preserve the area as $det(A^{T})=det(A)$. It means that the matrix
$B'=\begin{bmatrix}
a_{11} a_{22} - a_{21}^2 & a_{11} (a_{21} - a_{12})  \\
a_{22} (a_{12} - a_{21}) & a_{11} a_{22} - a_{12}^2 \\
\end{bmatrix}$
squares the area while transposing.
Edit 2:
The same matrix can be written as
$B'=\begin{bmatrix}
\begin{bmatrix}
a_{11} & a_{21} \\
\end{bmatrix}
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
\begin{bmatrix}
a_{11} & a_{21} \\
\end{bmatrix}
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
\begin{bmatrix}
a_{21} & a_{22} \\
\end{bmatrix}
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
\begin{bmatrix}
a_{12} & a_{22} \\
\end{bmatrix}
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
\end{bmatrix}$
Which is
$B'=\begin{bmatrix}
a_{1}^{T}
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
a_{1}^{T}
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
a_{2}^{T}
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
a_{2}^{T}
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
\end{bmatrix}=
\begin{bmatrix}
a_{1}\cdot
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
a_{1}\cdot
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
a_{2}\cdot
\begin{bmatrix}
a_{22} \\ -a_{21} \\
\end{bmatrix}
&
a_{2}\cdot
\begin{bmatrix}
-a_{12} \\ a_{11} \\
\end{bmatrix}
\\
\end{bmatrix}$
I find the vectors $c_{1}=\begin{bmatrix} a_{22} \\ -a_{21} \\ \end{bmatrix}$ and $c_{2}=\begin{bmatrix} -a_{12} \\ a_{11} \\ \end{bmatrix}$ interesting. When I draw them it looks like I only need to rotate each by 90 degress in different directions to end up with the transpose column vectors.
Edit 3:
Maybe I fool myself, but I think I'm getting closer. The column space
$C=
\begin{bmatrix}
c_{1}  & c_{2}  \\
\end{bmatrix}
=
\begin{bmatrix}
a_{22}  & -a_{12}  \\
-a_{21} & a_{11} \\
\end{bmatrix}$
is related to $A^{-1}$ because:
$AC=\begin{bmatrix}
a_{11}  & a_{12}  \\
a_{21} & a_{22} \\
\end{bmatrix}
\cdot
\begin{bmatrix}
a_{22}  & -a_{12}  \\
-a_{21} & a_{11} \\
\end{bmatrix}
=
\begin{bmatrix}
det(A)  & 0  \\
0 & det(A) \\
\end{bmatrix}
=det(A) I$
So
$C=A^{-1}det(A)$
B' can be written as well like this:
$B'=\begin{bmatrix}
\begin{bmatrix}
\begin{bmatrix}
a_{11}  & a_{12}  \\
a_{21} & a_{22} \\
\end{bmatrix}
&
\begin{bmatrix}
a_{22} \\
-a_{21} \\
\end{bmatrix}
\end{bmatrix}
\\
\begin{bmatrix}
\begin{bmatrix}
a_{11}  & a_{12}  \\
a_{21} & a_{22} \\
\end{bmatrix}
&
\begin{bmatrix}
-a_{12} \\
a_{11} \\
\end{bmatrix}
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
\begin{bmatrix}
\begin{bmatrix}
a_{11}  & a_{12}  \\
a_{21} & a_{22} \\
\end{bmatrix}
&
c_{1}
\end{bmatrix}
\\
\begin{bmatrix}
\begin{bmatrix}
a_{11}  & a_{12}  \\
a_{21} & a_{22} \\
\end{bmatrix}
&
c_{2}
\end{bmatrix}
\end{bmatrix}$
or like this
$B'=\begin{bmatrix}
\begin{bmatrix}
a_{11}  & a_{21}  \\
\end{bmatrix}
&
\begin{bmatrix}
a_{22}  & -a_{12}  \\
-a_{21} & a_{11} \\
\end{bmatrix}
\\
\begin{bmatrix}
a_{21}  & a_{22}  \\
\end{bmatrix}
&
\begin{bmatrix}
a_{22}  & -a_{12}  \\
-a_{21} & a_{11} \\
\end{bmatrix}
\\
\end{bmatrix}
=
\begin{bmatrix}
a_1^{T}
&
\begin{bmatrix}
c_{1}  & c_{2} \\
\end{bmatrix}
\\
a_2^{T}
&
\begin{bmatrix}
c_{1}  & c_{2} \\
\end{bmatrix}
\\
\end{bmatrix}
=
\begin{bmatrix}
a_1^{T}C
\\
a_2^{T}C
\\
\end{bmatrix}
=
det(A) \begin{bmatrix}
a_1^{T}A^{-1}
\\
a_2^{T}A^{-1}
\\
\end{bmatrix}$
Therefore for $BA=A^{T}$ we have
$B=\begin{bmatrix}
a_1^{T}A^{-1}
\\
a_2^{T}A^{-1}
\\
\end{bmatrix}$
Edit 4:
I think I will post my own answer soon. Going down the path of $A^{-1}$ had the idea that one can exploit the symmetry of of $AA^{T}$. Symmetry means that $AA^{T}$ decomposes nicer:
$AA^{T} = R_{AA^{T}} \Lambda_{AA^{T}} (R^{-1})_{AA^{T}}$
Now if you multiply both sides with $A^{-1}$ you'll get
$A^{T} = A^{-1} R_{AA^{T}} \Lambda_{AA^{T}} (R^{-1})_{AA^{T}}$
When I do an example with numbers I can also see that in my example $R_{AA^{T}} = (R^{-1})_{AA^{T}}$
$R_{AA^{T}}$ mirrors the space along y axis and then rotates by some angle $\alpha$ So my suspicion right now is:
$A^{T}=A^{-1} R_{AA^{T}} \Lambda_{AA^{T}} R_{AA^{T}}$
Now if I define
$R_{AA^{T}}^{'} = \begin{bmatrix}
cos \alpha & -sin \alpha \\
sin \alpha & cos \alpha \\
\end{bmatrix}$
to get the mirroring out of the matrix $R_{AA^{T}}$ then I get
$A^{T}=A^{-1}
R_{AA^{T}}^{'}
\begin{bmatrix}
-1 & 0 \\
0 & 1 \\
\end{bmatrix}
\Lambda_{AA^{T}}
R_{AA^{T}}^{'}
\begin{bmatrix}
-1 & 0 \\
0 & 1 \\
\end{bmatrix}
$
So generally
$A^{T}=A^{-1} R_{\alpha} M_y \Lambda R_{\alpha} M_y$
With $M_y$ being the mirroring along the y axis, $R_{\alpha}$ some counter-clockwise rotation by $\alpha$ and $\Lambda$ some scaling

REPLY [3 votes]: When trying to grasp the relation between  $A$, $A^T$ and $A^{-1}$, I created the attached plot
For $A^T$ this reads:

$\mathcal{r}_{U^T}$ is the rotation performed by $U^T$
$\mathcal{s}_{\Sigma}$ is the scaling performed by $\Sigma$
$\mathcal{r}_{V}$ is the rotation performed by $V$

The three axes show the SVD-decomposition of the three incarnations of $A$.

A green line between two axes indicates equality.
A red line indicates a contraposition.

In short, this says "$A^T$ scales like $A$, but rotates like $A^{-1}$."
So, $A^T$ has more in common with $A^{-1}$ then it has in common with $A$.
Not all matrices have an inverse.
If the inverse does not exist, the plot can still be made, replacing $A^{-1}$ with $A^{\dagger}$ and $\Sigma^{-1}$ with $\Sigma^{\dagger}$.
$A^{\dagger}$ is the generalized inverse of $A$.
Some more detail can be found on: www.heavisidesdinner.com
$A$, $A^T$ and $A^-1$">


TITLE: What is physics behind of explosion under Atmospheric pressure?
QUESTION [3 upvotes]: An explosion is a rapid increase in volume and release of energy in an extreme manner.
A blast wave in fluid dynamics is the pressure and flow resulting from the deposition of a large amount of energy in a small very localised volume.
The equation for a Friedlander waveform describes the pressure of the blast wave as a function of time:
$$P(t)=P_oe^{-\frac {t}{t^*}}(1-\frac {t}{t^*})$$
where P$_o$ is the peak pressure and t$^*$ is the time at which the pressure first crosses the horizontal axis (before the negative phase).
My Question:
Why does the pressure temporarily drop below ambient after the wave passes?

REPLY [0 votes]: Is this "vaccum" or "underpressure" vawe a pure mechanical thing that would occur also in a long air cylinder if we produce a shock wave with a plunger or piston?
In the case of an explosion I understand that shortly after the explosion the hot air bubble in the center starts to move upwards by convection, because it is less dense than the cooler aire around it. I understand that this produce a wind towards the explossion and possibly a "negative pressure" at some distance from the explosion near the ground, but I think it does not explain the underpresure spherical wave that seems to expand in 3D from the explosion (according to some videos in which a spherical condenstaion cloud is seen expanding spherically around the explosion in the first instants).


TITLE: $\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c} \ge \frac{16}{a+b+c+d}$ for $a,b,c,d>0$
QUESTION [1 upvotes]: Prove that
$$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c}
 \ge \frac{16}{a+b+c+d}$$
if $a,b,c,d>0$.
My attempt:
I've put this in the answers.
See also comments under this question.

REPLY [1 votes]: My attempt:
Let p=a+b+c+d then
$$
\frac{1}{b+c+d}+ \frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\equiv\\{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d}
$$
By the AM-GM inequality,
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq4\sqrt[4]{1\over(p-a)(p-b)(p-c)(p-c)}.
$$
Also by AM-GM inequality,
$$
\sqrt[4]{(p-a)(p-b)(p-c)(p-c)}\leq\frac14 \left((p-a)+(p-b)+(p-c)+(p-d)\right)\\
=\frac14(4p-p)=\frac34p\\
$$
so
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq {16\over 3p}
$$


text:
/-
Copyright (c) 2020 Scott Morrison. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Markus Himmel, Scott Morrison
-/
import category_theory.limits.shapes.zero_morphisms
import category_theory.limits.shapes.kernels
import category_theory.abelian.basic
import category_theory.subobject.lattice
import order.atoms

/-!
# Simple objects

We define simple objects in any category with zero morphisms.
A simple object is an object `Y` such that any monomorphism `f : X ⟶ Y`
is either an isomorphism or zero (but not both).

This is formalized as a `Prop` valued typeclass `simple X`.

In some contexts, especially representation theory, simple objects are called "irreducibles".

If a morphism `f` out of a simple object is nonzero and has a kernel, then that kernel is zero.
(We state this as `kernel.ι f = 0`, but should add `kernel f ≅ 0`.)

When the category is abelian, being simple is the same as being cosimple (although we do not
state a separate typeclass for this).
As a consequence, any nonzero epimorphism out of a simple object is an isomorphism,
and any nonzero morphism into a simple object has trivial cokernel.

We show that any simple object is indecomposable.
-/

noncomputable theory

open category_theory.limits

namespace category_theory

universes v u
variables {C : Type u} [category.{v} C]

section
variables [has_zero_morphisms C]

/-- An object is simple if monomorphisms into it are (exclusively) either isomorphisms or zero. -/
class simple (X : C) : Prop :=
(mono_is_iso_iff_nonzero : ∀ {Y : C} (f : Y ⟶ X) [mono f], is_iso f ↔ (f ≠ 0))

/-- A nonzero monomorphism to a simple object is an isomorphism. -/
lemma is_iso_of_mono_of_nonzero {X Y : C} [simple Y] {f : X ⟶ Y} [mono f] (w : f ≠ 0) :
  is_iso f :=
(simple.mono_is_iso_iff_nonzero f).mpr w

lemma simple.of_iso {X Y : C} [simple Y] (i : X ≅ Y) : simple X :=
{ mono_is_iso_iff_nonzero := λ Z f m, begin
    resetI,
    haveI : mono (f ≫ i.hom) := mono_comp _ _,
    split,
    { introsI h w,
      haveI j : is_iso (f ≫ i.hom), apply_instance,
      rw simple.mono_is_iso_iff_nonzero at j,
      unfreezingI { subst w, },
      simpa using j, },
    { intro h,
      haveI j : is_iso (f ≫ i.hom),
      { apply is_iso_of_mono_of_nonzero,
        intro w, apply h,
        simpa using (cancel_mono i.inv).2 w, },
      rw [←category.comp_id f, ←i.hom_inv_id, ←category.assoc],
      apply_instance, },
  end }

lemma kernel_zero_of_nonzero_from_simple
  {X Y : C} [simple X] {f : X ⟶ Y} [has_kernel f] (w : f ≠ 0) :
  kernel.ι f = 0 :=
begin
  classical,
  by_contra,
  haveI := is_iso_of_mono_of_nonzero h,
  exact w (eq_zero_of_epi_kernel f),
end

/--
A nonzero morphism `f` to a simple object is an epimorphism
(assuming `f` has an image, and `C` has equalizers).
-/
-- See also `mono_of_nonzero_from_simple`, which requires `preadditive C`.
lemma epi_of_nonzero_to_simple [has_equalizers C] {X Y : C} [simple Y]
  {f : X ⟶ Y} [has_image f] (w : f ≠ 0) : epi f :=
begin
  rw ←image.fac f,
  haveI : is_iso (image.ι f) := is_iso_of_mono_of_nonzero (λ h, w (eq_zero_of_image_eq_zero h)),
  apply epi_comp,
end

lemma mono_to_simple_zero_of_not_iso
  {X Y : C} [simple Y] {f : X ⟶ Y} [mono f] (w : is_iso f → false) : f = 0 :=
begin
  classical,
  by_contra,
  exact w (is_iso_of_mono_of_nonzero h)
end

lemma id_nonzero (X : C) [simple.{v} X] : 𝟙 X ≠ 0 :=
(simple.mono_is_iso_iff_nonzero (𝟙 X)).mp (by apply_instance)

instance (X : C) [simple.{v} X] : nontrivial (End X) :=
nontrivial_of_ne 1 0 (id_nonzero X)

section

lemma simple.not_is_zero (X : C) [simple X] : ¬ is_zero X :=
by simpa [limits.is_zero.iff_id_eq_zero] using id_nonzero X

variable [has_zero_object C]
open_locale zero_object

variables (C)

/-- We don't want the definition of 'simple' to include the zero object, so we check that here. -/
lemma zero_not_simple [simple (0 : C)] : false :=
(simple.mono_is_iso_iff_nonzero (0 : (0 : C) ⟶ (0 : C))).mp ⟨⟨0, by tidy⟩⟩ rfl

end
end

-- We next make the dual arguments, but for this we must be in an abelian category.
section abelian
variables [abelian C]

/-- In an abelian category, an object satisfying the dual of the definition of a simple object is
    simple. -/
lemma simple_of_cosimple (X : C) (h : ∀ {Z : C} (f : X ⟶ Z) [epi f], is_iso f ↔ (f ≠ 0)) :
  simple X :=
⟨λ Y f I,
 begin
  classical,
  fsplit,
  { introsI,
    have hx := cokernel.π_of_epi f,
    by_contra,
    substI h,
    exact (h _).mp (cokernel.π_of_zero _ _) hx },
  { intro hf,
    suffices : epi f,
    { exactI is_iso_of_mono_of_epi _ },
    apply preadditive.epi_of_cokernel_zero,
    by_contra h',
    exact cokernel_not_iso_of_nonzero hf ((h _).mpr h') }
 end⟩

/-- A nonzero epimorphism from a simple object is an isomorphism. -/
lemma is_iso_of_epi_of_nonzero {X Y : C} [simple X] {f : X ⟶ Y} [epi f] (w : f ≠ 0) :
  is_iso f :=
begin
  -- `f ≠ 0` means that `kernel.ι f` is not an iso, and hence zero, and hence `f` is a mono.
  haveI : mono f :=
    preadditive.mono_of_kernel_zero (mono_to_simple_zero_of_not_iso (kernel_not_iso_of_nonzero w)),
  exact is_iso_of_mono_of_epi f,
end

lemma cokernel_zero_of_nonzero_to_simple
  {X Y : C} [simple Y] {f : X ⟶ Y} (w : f ≠ 0) :
  cokernel.π f = 0 :=
begin
  classical,
  by_contradiction h,
  haveI := is_iso_of_epi_of_nonzero h,
  exact w (eq_zero_of_mono_cokernel f),
end

lemma epi_from_simple_zero_of_not_iso
  {X Y : C} [simple X] {f : X ⟶ Y} [epi f] (w : is_iso f → false) : f = 0 :=
begin
  classical,
  by_contra,
  exact w (is_iso_of_epi_of_nonzero h),
end

end abelian

section indecomposable
variables [preadditive C] [has_binary_biproducts C]

-- There are another three potential variations of this lemma,
-- but as any one suffices to prove `indecomposable_of_simple` we will not give them all.
lemma biprod.is_iso_inl_iff_is_zero (X Y : C) : is_iso (biprod.inl : X ⟶ X ⊞ Y) ↔ is_zero Y :=
begin
  rw [biprod.is_iso_inl_iff_id_eq_fst_comp_inl, ←biprod.total, add_right_eq_self],
  split,
  { intro h, replace h := h =≫ biprod.snd,
    simpa [←is_zero.iff_is_split_epi_eq_zero (biprod.snd : X ⊞ Y ⟶ Y)] using h, },
  { intro h, rw is_zero.iff_is_split_epi_eq_zero (biprod.snd : X ⊞ Y ⟶ Y) at h,
    rw [h, zero_comp], },
end

/-- Any simple object in a preadditive category is indecomposable. -/
lemma indecomposable_of_simple (X : C) [simple X] : indecomposable X :=
⟨simple.not_is_zero X,
λ Y Z i, begin
  refine or_iff_not_imp_left.mpr (λ h, _),
  rw is_zero.iff_is_split_mono_eq_zero (biprod.inl : Y ⟶ Y ⊞ Z) at h,
  change biprod.inl ≠ 0 at h,
  rw ←(simple.mono_is_iso_iff_nonzero biprod.inl) at h,
  { rwa biprod.is_iso_inl_iff_is_zero at h, },
  { exact simple.of_iso i.symm, },
  { apply_instance, },
end⟩

end indecomposable

section subobject
variables [has_zero_morphisms C] [has_zero_object C]

open_locale zero_object
open subobject

instance {X : C} [simple X] : nontrivial (subobject X) :=
nontrivial_of_not_is_zero (simple.not_is_zero X)

instance {X : C} [simple X] : is_simple_order (subobject X) :=
{ eq_bot_or_eq_top := begin
  rintro ⟨⟨⟨(Y : C), ⟨⟨⟩⟩, (f : Y ⟶ X)⟩, (m : mono f)⟩⟩, resetI,
  change mk f = ⊥ ∨ mk f = ⊤,
  by_cases h : f = 0,
  { exact or.inl (mk_eq_bot_iff_zero.mpr h), },
  { refine or.inr ((is_iso_iff_mk_eq_top _).mp ((simple.mono_is_iso_iff_nonzero f).mpr h)), }
end, }

/-- If `X` has subobject lattice `{⊥, ⊤}`, then `X` is simple. -/
lemma simple_of_is_simple_order_subobject (X : C) [is_simple_order (subobject X)] : simple X :=
begin
  split, introsI, split,
  { introI i,
    rw subobject.is_iso_iff_mk_eq_top at i,
    intro w,
    rw ←subobject.mk_eq_bot_iff_zero at w,
    exact is_simple_order.bot_ne_top (w.symm.trans i), },
  { intro i,
    rcases is_simple_order.eq_bot_or_eq_top (subobject.mk f) with h|h,
    { rw subobject.mk_eq_bot_iff_zero at h,
      exact false.elim (i h), },
    { exact (subobject.is_iso_iff_mk_eq_top _).mpr h, }, }
end

/-- `X` is simple iff it has subobject lattice `{⊥, ⊤}`. -/
lemma simple_iff_subobject_is_simple_order (X : C) : simple X ↔ is_simple_order (subobject X) :=
⟨by { introI h, apply_instance, },
 by { introI h, exact simple_of_is_simple_order_subobject X, }⟩

/-- A subobject is simple iff it is an atom in the subobject lattice. -/
lemma subobject_simple_iff_is_atom {X : C} (Y : subobject X) : simple (Y : C) ↔ is_atom Y :=
(simple_iff_subobject_is_simple_order _).trans
  ((order_iso.is_simple_order_iff (subobject_order_iso Y)).trans
    set.is_simple_order_Iic_iff_is_atom)

end subobject

end category_theory

subset_name: curated


text:
\begin{document}
\begin{abstract}
  We obtain sharp asymptotics for the probability that the
  $(2+1)$-dimensional discrete SOS interface at low temperature is
  positive in a large region. For a square region $\L$, both
  under the infinite volume measure and under the measure with zero
  boundary conditions around $\L$, this  probability turns out to behave
  like $\exp(-\tau_\beta(0) L \log L )$, with $\tau_\beta(0)$ the
  surface tension at zero tilt, also called step free energy, and $L$ the box side. This behavior is qualitatively different from
  the one found for continuous height massless gradient
  interface models \cite{BDZ,DG}.
\end{abstract}

\keywords{SOS model, Loop ensembles, Random surface models, Entropic repulsion, Large deviations.}
\subjclass[2010]{60K35, 60F10, 82B41, 82C24}

\maketitle

\section{Introduction}

Let $\bbP_\L$ denote the Gibbs measure of the $(2+1)$-dimensional SOS
model on a box $\L\subset \mathbb Z^2$ with zero boundary condition. The configurations
are discrete height functions $\eta:\Lambda\mapsto
\mathbb Z$ whereas
$\eta(x)=0$ for $x\notin \L$.
The probability
measure is given by
\[
\bbP_\L(\eta)=\frac{\exp{\big(-\beta\sum_{|x-y|=1}|\eta(x)-\eta(y)|\big)}}{Z_\L},
\]
where $\beta>0$ is the inverse temperature, and $Z_\L$ denotes
the associated normalizing factor, called partition function. We will mostly consider the case where
 $\L=\L_L=[-L,L]^2\cap\bbZ^2$ is the square of side $2L+1$ in $\bbZ^2$ centered at the origin.

It is well known that, if $\beta$ is
sufficiently large (as we assume from here on), the
limit of $\bbP_{\L_L}$ as $L\to\infty$ exists (in the sense that
the probability of any local event converges), and is denoted
$\bbP$, the infinite-volume Gibbs measure; see e.g.\ \cite{BW}.

The infinite volume measure is characterized by the fact that heights
have finite variance and exponentially decaying tails: the interface
is globally very rigid and flat, the height is exactly
zero on a set of sites of density $1-O(\exp(-4\beta))$ and typical fluctuations are isolated
spikes; see \cite{BW,Bricetal,CLMST2}. The question we investigate here is that of large fluctuations
of the interface, namely, the asymptotics of the probability that the
interface is positive in a fixed large region.
In order to formulate our main result, let us recall the definition of
the {\em surface tension} at zero tilt, often referred to as {\em step free energy}:
\begin{definition}\label{deftb}
Let $\xi$ be the height function on $\L_L^c$ such that
  $\xi(x)=1$ if $x=(x_1,x_2)$ with
$x_2\geq0$,  and $\xi(x)=0$ otherwise.
Let $Z_{\L_L}^{\xi}$ be the
partition function on $\L_L$ with boundary condition $\xi$ (see Section \ref{notation} below for more details).
Then, the  surface tension at zero tilt is defined as
\begin{eqnarray}
  \label{eq:3}
  \tau_\b(0)=-\lim_{L\to\infty}\frac{1}{2\beta L}\log\frac{ Z_{\L_L}^{\xi}}{ Z_{\L_L}}.
\end{eqnarray}
\end{definition}
It is a known fact that $\t_\b(0)$ is well defined and that, for $\b$ sufficiently large, one has $\t_\b(0)>0$, see Lemma
\ref{lem:surftens} below for more details.
We have then:

\begin{theorem}\label{th:main}
There exists $\b_0>0$ such that for any $\b\geq \b_0$ one has
\begin{align}\label{LDsos}
\lim_{L\to\infty}\frac1{L\log L}\log \bbP_{\L_L} \left( \eta(x) \geq
0\;\text{for every\;} x\in \L_L\right) = -2\t_\b(0).
\end{align}
The same limit holds if we replace $\bbP_{\L_L}$ by $\bbP$.
\end{theorem}
Actually, it will be clear from the proof
that the result still holds if we
replace the inequality $\eta(x) \geq
0 $ with $\eta(x)\geq n$, for any fixed $n>0$.

\smallskip

We now describe the heuristics behind Theorem \ref{th:main}. In
\cite{CLMST2} (see also \cite{CRAS} for a summary of the main results) the scaling limit of the shape of the SOS
surface in the box $\L_L$ with zero boundary conditions and
conditioned  to be non-negative was established in full detail.
The SOS
interface lifts rigidly to a height
$H(L)=\lfloor\tfrac1{4\beta}\log L\rfloor$, in order to create room
for downward spike-like fluctuations
(entropic repulsion). As a consequence there are $H(L)$ macroscopic level lines,
following approximately $\partial\L_L$, where the height of
the surface jumps (roughly) by one. A fraction $1-o(1)$ of
the level lines is at distance $o(L)$ from $\partial \L_L$ while the rest
has a non trivial scaling limit as $L\to \infty$, with flat and curved
parts and $1/3$
fluctuation exponent along the flat part.
Roughly each of the level lines at distance $o(L)$ from $\partial\L_L$ entails
a surface energy cost $|\partial \L_L|\beta\tau_\b(0)=8\beta L \tau_\b(0)$. The
total energy cost of the macroscopic level lines ensemble is therefore
\[
(1-o(1))8\beta \tau_\b(0)H(L) L=2(1-o(1)) \tau_\b(0) L \log L,
\]
which explains \eqref{LDsos}.
The difficulty that arises in substantiating this heuristics is that
the $H(L)$ contours have mutual interactions. If these are naively
estimated, they produce an additive term, of apriori indefinite sign, of order $O(c_\beta
|\partial\L_L|H(L))=O(\gep_\beta L\log L)$ in the energy
cost. Here $\gep_\beta=c_\b/\b>0$ is a constant tending to zero as
$\beta\to\infty$, but non-zero for any finite $\beta$. While this
problem can be avoided when
looking for   a lower bound on the l.h.s. of \eqref{LDsos},  simply by
imposing that the contours stay sufficiently far one from the other to
neglect the interaction, as an upper
bound we would get nothing better than $-2\tau_\b(0)+\gep_\beta$.

The solution we
find is an iterative monotonicity argument (Theorem \ref{th:m1}), based on the FKG
properties of the SOS model, which we
believe is of interest by itself. This allows us to conclude that the
possibly attractive effect of the mutual interaction potential is more
than compensated by the loss of entropy due to the fact that the
contours cannot mutually cross. As a consequence, the surface tension
associated to $n$ SOS contours is at least the sum of the individual
surface tensions (Corollary \ref{coro_surftens})\footnote{After completing this work we realized that a conceptually similar argument was put forward by Bricmont, El Mellouki and Fr\"ohlich \cite[Appendix 1]{Bricetal} to compare the step free energy to the free energy associated to a single macroscopic step in the boundary condition.}.

\subsection{Discussion} Since the early work of Lebowitz and Maes
\cite{LebMaes}, the problem of computing the sharp large deviation
behavior of the positivity event $\eta(x)\geq 0$, $x\in\L_L$, has
attracted much attention. Refined estimates have been obtained for
continuous height models such as the Gaussian free field on $\bbZ^d$,
see \cite{BDZ,BDG,Dzero}, as well as for more general lattice massless
free fields \cite{DG}. A large deviation theory for such models was
further developed in \cite{DGI}. The problem is of particular
relevance in the study of the entropic repulsion phenomenon
\cite{Bricetal}, see e.g.\ \cite{Yvan_survey} for a survey.
Considerable progress has been recently made for the SOS model
\cite{CLMST,CRAS,CLMST2} and for the discrete Gaussian model
\cite{LMS} for which the SOS gradient term $|\eta(x)-\eta(y)|$ in the
energy function is replaced by
$(\eta(x)-\eta(y))^2$, but the question of computing the limit in \eqref{LDsos} remained unaddressed.

As a matter of comparison, let us briefly recall the known results for the two-dimensional continuous Gaussian case.

If $\bbP_L$ denotes the
$2$D Gaussian free field on $\L_L$ with zero boundary condition, then for any $\d\in(0,1)$
one has
\begin{align}\label{LDGFF}
\lim_{L\to\infty}\frac1{(\log L)^2}\log \bbP_{L} \left( \eta(x) \geq
0\;\text{for every\;} x\in \L_{(1-\d)L}\right) = -\kappa(\d),
\end{align}
where $\kappa(\d)>0$ is a constant related to the relative capacity of the set $\L_{(1-\d)L}$ with respect to  $\L_L$ which satisfies $\kappa(\d)\to\infty$ as $\d\to 0$; see \cite[Theorem 3]{BDG}. On the other hand, boundary effects  dominate if {\em all} heights in $\L_L$ are required to be nonnegative, and one expects \cite[Section 3]{DG}
that
 \begin{align}\label{LDGFF0}
\lim_{L\to\infty}\frac1{L}\log \bbP_{L} \left( \eta(x) \geq
0\;\text{for every\;} x\in \L_{L}\right) = -\chi,
\end{align}
for some $\chi>0$.
Because of its discrete nature, the SOS interface considered in our work presents a very different behavior.
First, the rigidity of the interface allows one to consider the infinite volume limit - whereas the $2$D massless free field does not admit such a limit. Second, while the typical height in the bulk under the positivity constraint is of order $\log L$ just as in the case of the $2$D massless free field,
the cost of such a shift is much higher due to the unavoidable presence of as many as $H(L)$ macroscopic level lines each of which has a definite cost proportional to the length. In particular, boundary terms do not dominate here and the estimate of Theorem \ref{th:main} holds for $\bbP$ as well as for $\bbP_{\L_L}$.

\section{Contours, surface tension, etc.}\label{tools}
Here we define the model, and the notion of contours of the SOS interface.
To express the law of contours we shall use a {\em cluster expansion} for partition functions of the SOS
model. Finally we recall the definition of surface tension for a
general tilt, and some of its properties.

\subsection{SOS model: basic definitions and notation}\label{notation}
We call a {\em bond} (resp. {\em
  dual bond})
any straight line segment joining two neighboring sites in ${\bbZ^2}$ (resp. of
${\bbZ^2}^*$, the dual lattice of $\bbZ^2$). Here $\bbZ^2$ and ${\bbZ^2}^*\equiv \bbZ^2+(1/2,1/2)$ are thought of as embedded in $\bbR^2$.
For any finite $\L\subset\bbZ^2$, let $\cB_\L\subset {\bbZ^2}$ denote
the set of bonds of the form $e=xy$  with $x\in\L$ and $y \in
\L\cup \partial \L$, where $\partial \L$ is the external boundary of
$\L$, i.e.\ the set of $y\in\L^c$ such that $xy$ is a bond for some $x\in\L$.
A height configuration $\t: \L^c\mapsto \bbZ$ is called a boundary condition.
We define $\O_\L^\t$ as the set of height functions $\eta:\bbZ^2\mapsto \bbZ$ such that $\eta(x)=\t(x)$ for all $x\notin \L$.
The SOS Hamiltonian in $\L$ with boundary condition $\t$ is the function defined by
\begin{align}\label{sos_ham}
\cH_\L^\t (\eta) = \sum_{xy\in\cB_\L} |\eta(x)-\eta(y)| \,,\quad \;\eta\in\O_\L^\t.
\end{align}
The SOS Gibbs measure in $\L$ with  boundary condition $\t$ at inverse temperature $\b$ is the probability measure
$\bbP_\L^\t$ on $\O_\L^\t$ given by
\begin{align}\label{sos_gibbs}
\bbP_\L^\t(\eta) = \frac1{Z_\L^\t}\,\exp{\left(-\b\cH_\L^\t (\eta)\right)}\,,
\end{align}
where $Z_\L^\t$ is the partition function $$Z_\L^\t=\sum_{\eta\in\O_\L^\t}\exp{\left(-\b\cH_\L^\t (\eta)\right)}.$$ When $\t=0$ we simply write $Z_\L$ for $Z_\L^0$ and $\bbP_\L$ for $\bbP_\L^0$.
We often consider boxes $\L$ of rectangular shape, and write $\L_{L,M}$, with $L,M\in \bbN$, for the rectangle
$\L_{L,M}=([-L,L]\times [-M,M])\cap\bbZ^2$ centered at the origin. When $L=M$ we  write $\L_L$ for the square of side $2L+1$.

We recall that the SOS model satisfies the so called FKG
inequality~\cite{FKG} with respect to the natural partial order defined by
$\eta\leq\eta'\Leftrightarrow \eta(x)\leq \eta'(x)$ for every $x$.
That is,
if $f$ and $g$ are two increasing (w.r.t.\ the above partial order)
functions, then
$\bbE^\t_\Lambda(fg)\ge \bbE^\t_\Lambda(f) \bbE^\t_\Lambda (g)$ for any region $\L$ and any boundary condition $\t$, where $\bbE^\t_\L$ denotes expectation w.r.t.\ $\bbP_\L^\t$.
To prove the FKG inequality one can establish directly the validity of the FKG lattice condition
\begin{equation}\label{holley}
\bbP_\L^{\t}(\eta \vee \eta') \bbP_\L^{\t}(\eta\wedge \eta')\ge \bbP_\L^{\t}(\eta) \bbP_\L^{\t}(\eta').
\end{equation}

\subsection{Geometric contours, $h$-contours etc.}\label{hcont}
We use the following notion of contours.
\begin{definition}\label{contourdef}
Two sites $x,y$ in $\bbZ^2$ are said to be {\em separated by a dual bond $e$} if
their distance (in $\bbR^2$) from $e$ is $\tfrac12$. A pair of
orthogonal dual
bonds which meet in a site $x^*\in {\Z^2}^*$ is said to be a
{\sl linked pair of bonds} if both are on the same side of the
forty-five degrees line (w.r.t. to the horizontal axis) across $x^*$. A {\sl geometric contour} (for
short a contour in the sequel) is a
sequence $e_0,\ldots,e_n$ of dual bonds such that:
\begin{enumerate}
\item $e_i\ne e_j$ for $i\ne j$, except for $i=0$ and $j=n$ where $e_0=e_n$.
\item for every $i$, $e_i$ and $e_{i+1}$ have a common vertex in ${\Z^2}^*$.
\item if $e_i,e_{i+1},e_j,e_{j+1}$ all have a common vertex $x^*\in {\Z^2}^*$,
then $e_i,e_{i+1}$ and $e_j,e_{j+1}$ are linked pairs of bonds.
\end{enumerate}
We denote the length of a contour $\gamma$, i.e.  the number of distinct bonds in $\g$, by $|\gamma|$, its interior
(the sites in $\bbZ^2$ it surrounds) by $\Lambda_\gamma$ and its
interior area (the number of such sites) by
$|\Lambda_\gamma|$. Moreover we let $\Delta_{\gamma}$ be the set of sites in $\Z^2$ such that either their distance
(in $\bbR^2$) from $\gamma$ is $\tfrac12$, or their distance from the set
of vertices in ${\Z^2}^*$ where two non-linked bonds of $\gamma$ meet
equals $1/\sqrt2$. Finally we let $\Delta^+_\gamma=\Delta_\gamma\cap
\L_\g$ and $\Delta^-_\gamma = \Delta_\gamma\setminus \Delta^+_\gamma$.
Given a contour $\gamma$ we say that $\gamma$ is an \emph{$h$-contour}
for the configuration $\eta$ if
\[
\eta\restriction_{\Delta^-_\gamma}\leq h-1, \quad \eta\restriction_{\Delta^+_\gamma}\geq h.
\]

Finally $\sC_{\gamma,h}$ will denote the event that $\gamma$ is an $h$-contour.
\end{definition}

To illustrate the above definitions with a simple example, consider the elementary contour given by the square of side $1$ surrounding a site $x\in\bbZ^2$. In this case, $\g$ is an $h$-contour iff $\eta({x})\geq h$ and $\eta(y)\leq h-1$ for all $y\in\{x\pm e_1, x\pm e_2, x+ e_1+e_2,x-e_1-e_2\}$.

We observe that
a geometric contour $\g$ could be at the same time a
$h$-contour and a $h'$-contour with $h\neq h'$. More generally two
geometric contours $\g,\g'$ could be contours for the same surface
with different height parameters even if $\g\cap\g'\neq\emptyset$, but then
the interior of one one of them must be contained in the interior of the other; see Figure \ref{fig:002} for an example.

\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.37]{contours3}

\end{overpic}

        \caption{
        Example of a SOS configuration above the wall in the $7\times 7$ box $\L_3$ with zero boundary conditions: white sites have height $0$, shaded sites have height $1$ and darker sites have height $2$. Notice that according to Definition \ref{contourdef} there are three $1$-contours and two  $2$-contours.
          }\label{fig:002}
\end{figure}

\subsection{Cluster expansion}\label{cl_exp}
So called cluster expansions are a well established tool for the analysis of random interfaces at low-temperature; see e.g.\ \cite{BW} where both the SOS model and the discrete gaussian model are considered. Here we shall need a particular expansion that allows us to take into account the extra constraints that appear naturally
in the partition function of the SOS model on a region $\L$ delimited by two contours; see
   \eqref{ap20} below.

Given a finite connected set $\L\subset \bbZ^2$, let
$\partial _*\L$ denote the set of  $y\in \L$ either at distance $1$ from $\partial \L$
or at distance $\sqrt 2$ from $\partial \L$ in the south-west or north-east direction.
Fix $U_+,U_-\subset \partial_* \L$, and let
$ Z_{\L,U_+,U_-}$ denote the SOS partition function in $\L$ with the sum over $\eta$  restricted to those
$\eta\in \O_\L^0$ such that $\eta(x)\geq 0$ for all $x\in U_+$ and $\eta(x)\leq 0$ for all $x\in U_-$.
Clearly, if $U_-\cap U_+\neq \emptyset$, then $\eta(x)=0$ is fixed for
all $x\in U_-\cap U_+$. If $\L=\emptyset$ then $Z_{\L,U_+,U_-}:= 1$.
We refer the reader to \cite[App. A]{CLMST} for a proof of the following statement based on the general cluster expansion from \cite{KotPre}.
\begin{lemma}
\label{lem:dks}
There exists $\b_0>0$ independent of $\L$ such that for all $\b\geq \b_0$, for all finite connected $\L\subset\bbZ^2$ and $U_+,U_-\subset \partial_* \L$:
\begin{equation}\label{zetaexp}
\log  Z_{\L,U_+,U_-}=\sum_{V\subset \L}\varphi_{U_+,U_-}(V),
\end{equation}
where the potentials $\varphi_{U_+,U_-}(V)$ satisfy
\begin{enumerate}[(i)]
\item $\varphi_{U_+,U_-}(V)=0$ if $V$ is not connected.
\item $\varphi_{U_+,U_-}(V)=\varphi_0(V)$ if $V\cap({U_+\cup U_-})=\emptyset$, for
  some shift invariant  potential $V\mapsto \varphi_0(V)$,

that is
\[
\varphi_0(V)=
\varphi_0(V+x)\quad \forall \,x\in\bbZ^2,
\]
where $\varphi_0$ is independent of $U_+,U_-,\L$.
\item For all $V\subset \L$:
\[

\sup_{{U_+,U_-}\subset\partial_* \L}|\varphi_{U_+,U_-}(V)|\leq
  \exp(-(\b-\b_0)\, d(V))
\]
where $d(V)$ is the cardinality of the smallest connected set of all dual
bonds separating points of $V$ from points of its
complement (a dual bond separates $V$ from $V^c$ iff it is orthogonal to a bond connecting $V$ to $V^c$).
\end{enumerate}
\end{lemma}

\subsection{Nested contours}\label{wedding}
Consider the rectangle $\L_{L,M}$,  for some $L,M\in \bbN$,
and let $\bbP_{\L}$ denote the $SOS$ Gibbs measure in $\L:=\L_{L,M}$ with zero boundary conditions. Given two contours $\g,\g'$, we write $\g\subset \g'$ if $\L_\g\subset\L_{\g'}$. Fix $n\in\bbN$ and pick $n$ geometric contours $\g_1,\dots,\g_n$ such that
$ \g_{i+1}\subset \g_i$, for every $i=1,\dots,n-1$.
Consider the event $\cap_{i=1}^n\sC_{\g_i,i}$ that $\g_i$ is an
$i$-contour for all $i=1,\dots,n$. The probability of this event under $\bbP_{\L}$ can be expressed as
\begin{align}\label{wedprob}
\bbP_{\L}\big(\cap_{i=1}^n\sC_{\g_i,i}\big) = \frac{Z(\g_1,\dots,\g_n; L,M)}{Z_\L},
\end{align}
where $Z_\L$ denotes the partition function of the $SOS$ model in $\L=\L_{L,M}$ with zero boundary conditions and $ Z(\g_1,\dots,\g_n; L,M)$ stands for the same summation restricted to the configurations $\eta\in\O_\L^0$ such that $\g_i$ is an $i$-contour for each $i=1,\dots,n$.
By applying the  cluster expansion in Lemma \ref{lem:dks}, with $\L=\L_{L,M}$ and $U_\pm=\emptyset$, we can write
\begin{align}\label{ub202}
Z_\L = \textstyle \exp{\big(\sum_{V\subset\L}\varphi_0(V)\big)}.
\end{align}
To expand the partition function $Z(\g_1,\dots,\g_n; L,M)$, define $S_i:=\L_{\g_{i-1}}\setminus \L_{\g_i}$, for $i=1,\dots,n+1$, where $\L_{\g_{0}}=\L$ and $\L_{\g_{n+1}} =\emptyset$, and set
$\D_i^+=S_i\cap \D^+_{\g_{i-1}}$, and $\D_i^-=S_i\cap \D^-_{\g_{i}}$, with the understanding that $\D_1^+ = \D_{n+1}^- =\emptyset$. Notice that $\D_i^\pm\subset \partial_*S_i$.
Using the notation of Lemma \ref{lem:dks} a direct computation proves that
\begin{align}\label{ap20}
Z(\g_1,\dots,\g_n; L,M)= \textstyle \exp{\big(-\b\sum_{i=1}^n|\g_i|\big)}\prod_{i=1}^{n+1}Z_{S_i,\D_i^+,\D_i^-}\,.
\end{align}
The term $\sum_{i=1}^n|\g_i|$
accounts for the minimal energy associated to the given contours. The
fact that the surface gradient across a contour $\g_i$
must be at least $1$ is encoded by the constraints on $\D_i^+,\D_i^-$
appearing in $Z_{S_i,\D_i^+,\D_i^-}$.

Therefore, the expansion \eqref{zetaexp} implies
\begin{align}\label{ap200}
Z(\g_1,\dots,\g_n; L,M)= \textstyle \exp{\big(-\b\sum_{i=1}^{n}|\g_i|+ \sum_{i=1}^{n+1}\sum_{V\subset S_i}\varphi_{\D_i^+,\D_i^-}(V)\big)}.
\end{align}
The ratio \eqref{wedprob} then becomes
\begin{align}\label{ub2a1}
\bbP_{\L}\big(\cap_{i=1}^n\sC_{\g_i,i}\big) =
\textstyle \exp{\big(-\b\sum_{i=1}^n|\g_i| +
 \Psi_\L(\g_1,\dots,\g_n)\big)},
\end{align}
where
\begin{align}\label{ub211}
\Psi_\L(\g_1,\dots,\g_n) =
\sum_{i=1}^{n+1}\!\!\!\!\sumtwo{V\subset S_i:}{ V\cap (\D_i^+\cup\D_i^-)  \neq \emptyset}\!\!\!\!\!\!(\varphi_{\D_i^+,\D_i^-}(V)-\varphi_0(V)) -\!\!\!\! \sumtwo{V\subset\L:}{V\cap (\cup_{i=1}^n\g_i) \neq \emptyset} \varphi_0(V),
\end{align}
where the condition $V\cap (\cup_{i=1}^n\g_i) \neq \emptyset$ means that $V$ intersects more than just one $S_i$.
When $n=1$, we have only one contour $\g_1=\g$ and we define
\begin{align}\label{ub1c}
\psi_\L(\g):=\Psi_\L(\g_1) .
\end{align}
Observe that property (iii) of the potentials $\varphi_{U_+,U_-}(V)$ in Lemma \ref{lem:dks} implies in particular that
\begin{equation}
\label{rev1}
|\psi_\L(\g)|\leq 3\sum_{V\cap(\D^+_\g\cup\D^-_\g)\neq \emptyset}e^{-2(\b-\b_0)d(V)} \leq \e(\beta)|\g|,
\end{equation}
where $\lim_{\b\to \infty}\e(\b)=0$. Later on it will be convenient to introduce the quantity $\psi_\infty(\g)$
defined as $\psi_\L(\g)$ but without the restriction $V\subset \L$, i.e. now $S_1= \bbZ^d\setminus\L_\g$.
\subsection{The staircase ensemble}\label{stair}
Consider the rectangle $\L_{L,M}$, for some $L,M\in \bbN$.
Fix $n\in\bbN$ and integers
\begin{equation}
\label{ref2}
-M\leq a_1\leq \cdots\leq a_n\leq M,\;\;\text{ and}\;\; -M\leq b_1\leq \cdots\leq b_n\leq M,
\end{equation}
and set $a_0=b_0=-(M+1)$ and $a_{n+1}=b_{n+1}=M+1$. We define a ``staircase'' height $\t$ at the external boundary $\partial \L_{L,M}$ of our rectangle which, starting from
height zero at the base of the rectangle (i.e. the set $(x,-(M+1)), x=-L,\dots,L$)  jumps by one at the locations specified by the two
$n$-tuples $\{a_i,b_i\}$ until it reaches height $n$:
\begin{align}\label{stairtau}
\tau(u,v)=
  \begin{cases}
  i & \ \text{if $u=-L-1$ and $a_i \leq v<a_{i+1}$ or $u=L+1$ and $b_i \leq v<b_{i+1}$, }\\
 0 & \  \text{if $u\in[-L,L]$ and $v=-M-1$}\\
n & \ \text{if $u\in[-L,L]$ and $v=M+1$},
  \end{cases}
\end{align}
where $i\in\{0,\dots,n\}$, see Figure \ref{fig:1}.
Note that if two or more values of the $a_i$ or $b_i$ coincide then the boundary height $\t$ takes jumps
higher than $1$ at those points.

Next, let $Z(a_1,\dots,a_n; b_1,\dots, b_n; L,M)$ denote the partition function of the $SOS$ model in $\L_{L,M}$ with boundary condition $\t$ as in \eqref{stairtau}.
\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.27]{Fig1}

\put(30.5,29){$z_1$}\put(30.5,31){$\cdot$}
\put(47,33.5){$z_2$}\put(47,35.8){$\cdot$}
\put(56,17){$z'_1$}
\put(68,21){$z'_2$}
\put(68.5,23.7){$\cdot$}
\put(56,19.5){$\cdot$}
\end{overpic}

        \caption{A sketch of the staircase boundary condition (\ref{stairtau}) in the rectangle $\L_{L,M}$
          for $n=2$. The points $z_i,z'_i$ have coordinates $z_i=(-L-1,a_i)$, and  $z'_i=(L+1,b_i)$.}
          \label{fig:1}
\end{figure}

Let also $Z_\L$ denote as above the partition function of the $SOS$ model in $\L=\L_{L,M}$ with zero boundary condition everywhere. We want to compute the ratio
\begin{align}\label{rat}
\frac{Z(a_1,\dots,a_n; b_1,\dots, b_n; L,M)}{Z_\L}.
\end{align}
To expand the partition function in the numerator of \eqref{rat}, we
need the notion of an {\em open contour}. This is defined as in
Definition \ref{contourdef} except that $e_0\neq e_n$. Since the boundary conditions force the surface height to grow from height zero at the bottom base of $\L_{L,M}$ to height $n$ at the top base, necessarily any configuration $\eta$ of the $SOS$ interface compatible with $\tau$ must satisfy the following property.

Given $\eta$ there exist uniquely defined non-crossing open contours
$\g_i$, $i=1,\dots, n$, joining the dual lattice points
$x_i:=(-L-1/2,a_i-1/2)$ and $ y_i:=(L+1/2,b_i-1/2)$ such that
$\eta(x)\leq i-1$ for all $x\in\D_i^-$ and $\eta(x)\geq i-1$ for all
$x\in\D_i^+$ where $\D_i^\pm$ are now the sets defined as follows. Let
$S_i\subset \L_{L,M}$ denote the region bounded by $\g_i$ and
$\g_{i-1}$, where $\g_{n+1}$ is the top boundary of $\L_{L,M}$ and
$ \g_{0}$ is the bottom boundary of $\L_{L,M}$.  Then $\D_i^-$ is
defined as the set of $x\in S_i$ such that either their distance from
$\g_{i}$ is $\tfrac12$, or their distance from the set of vertices in
${\Z^2}^*$ where two non-linked bonds of $\g_{i}$ meet equals
$1/\sqrt2$. Similarly, $\D_i^+$ is the set of $x\in S_i$ such that
either their distance from $\g_{i-1}$ is $\tfrac12$, or their distance
from the set of vertices in ${\Z^2}^*$ where two non-linked bonds of
$\g_{i-1}$ meet equals $1/\sqrt2$. Lemma \ref{lem:dks}
here implies
\begin{align}\label{app200}
&Z(a_1,\dots,a_n; b_1,\dots, b_n; L,M)  \nonumber
\\
 &\qquad=\textstyle \sum_{\g_1,\dots,\g_n}\exp{\big(-\b\sum_{i=1}^{n}|\g_i|+ \sum_{i=1}^{n+1}\sum_{V\subset S_i}\varphi_{\D_i^+,\D_i^-}(V)\big)},
\end{align}
where the sum ranges over all possible values of the open contours $\g_i:x_i\to y_i$ inside $\L_{L,M}$ with the non-crossing constraints.
Recalling that $Z_\L$ can be expanded as in \eqref{ub202}, one finds that
\begin{align}\label{ub3a1}
 &\frac{Z(a_1,\dots,a_n; b_1,\dots, b_n; L,M)}{Z_\L}\nonumber
\\
 &\qquad=
\textstyle\sum_{\g_1,\dots,\g_n}\exp{\big(-\b\sum_{i=1}^n|\g_i| +
 \Phi_{L,M}(\g_1,\dots,\g_n)\big)},
\end{align}
where
\begin{align}\label{ub311}
\Phi_{L,M}(\g_1,\dots,\g_n) =
\sum_{i=1}^{n+1}\!\!\!\!\sumtwo{V\subset S_i:}{ V\cap (\D_i^+\cup\D_i^-)  \neq \emptyset}\!\!\!\!\!\!(\varphi_{\D_i^+,\D_i^-}(V)-\varphi_0(V)) - \!\!\!\!\!\!\!\!\sumtwo{V\subset \L_{L,M}:}{V\cap (\cup_{i=1}^n\g_i) \neq \emptyset} \varphi_0(V),
\end{align}
where the condition $V\cap (\cup_{i=1}^N\g_i) \neq \emptyset$ meansthat $V$ intersects more than just one $S_i$.
Equation \eqref{ub3a1} expresses the ratio \eqref{rat} as the partition function of a gas of
$n$ interacting non-crossing open contours $\g_1,\dots,\g_n$ within $\L_{L,M}$
such that $\g_i:x_i\to y_i$, $i=1,\dots,n$.
 Using (iii) in Lemma \ref{lem:dks} the limit as $M\to\infty$ of the above expression is well defined, so that the following holds.
 \begin{lemma}\label{lem:infvol}
For any integers $n$, $\{a_i,b_i\}_{i=1}^n$ satisfying \eqref{ref2}, the limit   \begin{equation}
    \label{eq:1}
\cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\,
  L) :=    \lim_{M\to \infty}\frac{Z(a_1,\dots,a_n;\, b_1,\dots,b_n;\,
  L,M)}{Z_\L}
  \end{equation}
 exists and it satisfies
 \begin{align}\label{ub4a1}
 &\cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\,
  L) = \nonumber
\\
 &\qquad=
\textstyle\sum_{\g_1,\dots,\g_n}\exp{\big(-\b\sum_{i=1}^n|\g_i| +
 \Phi_{L,\infty}(\g_1,\dots,\g_n)\big)},
\end{align}
where the sum ranges over all possible values of the open contours in the strip $\L_{L,\infty}:=[-L,L]\times \bbZ$
and $\Phi_{L,\infty}$ is defined as in \eqref{ub311} with $\L_{L,M}$ replaced by $\L_{L,\infty}$.
 \end{lemma}
 \begin{proof}
  Using (iii) in Lemma \ref{lem:dks} it is immediate to check that for any family of contours $(\g_1,\dots,\g_n)$
\[
  \lim_{M\to \infty} \Phi_{L,M}(\g_1,\dots,\g_n)= \Phi_{L,\infty}(\g_1,\dots,\g_n).
\]
As in \eqref{rev1} we have
\begin{equation}
\label{rev4}
|\Phi_{L,M}(\g_1,\dots,\g_n) |\leq \e(\beta)\sum_{i=1}^n|\g_i|.
\end{equation}
Hence, for $\beta $ large enough, the conclusion follows by dominated convergence.
 \end{proof}
\subsection{Surface tension}\label{surftens}
Here we recall the definition and some properties of the surface tension corresponding to arbitrary tilt.
Let us rewrite \eqref{ub4a1} in the case $n=1$ as
\begin{align}\label{contourmod}
 &\cZ(a_1;b_1;
  L) =
\sum_{\g}\exp{\big(-\b|\g| +
 \Phi_{L,\infty}(\g)\big)},
\end{align}
where the sum ranges over all open contours in the strip $\L_{L,\infty}$
joining the dual lattice points
$x_1:=(-L-1/2,a_1-1/2)$ and $ y_1:=(L+1/2,b_1-1/2)$.

\begin{lemma}\label{lem:surftens}
There exists $\b_0>0$ such that the following holds for all $\b\geq \b_0$. Let $\cZ(a_1;\, b_1;\,   L)$, denote the partition function \eqref{contourmod}.
Assume that as $L\to\infty$ one has $(b_1-a_1)/(2L)\to \l\in\bbR$ and set $\theta=\tan^{-1}(\l)$.   Then the function
\begin{equation}
    \label{eq:surftens1}
\t_\b(\theta)=-\lim_{L\to \infty}\frac{\cos(\theta)}{2\b L}\log \cZ(a_1;\,b_1;\,   L),
  \end{equation}
 is well defined and positive in $(-\pi/2,\pi/2)$. It is convex in the following
 sense: defining, for $x\in\mathbb R^2$,
 $\tau_\beta(x)=\|x\|\tau_\beta(\theta_x)$ with $\theta_x$ the  angle
 formed by the vector $x$ with the horizontal axis, $\tau_\beta$ is a
 convex function on $\mathbb R^2$.
 Moreover,
 \begin{equation}
    \label{eq:surftens2}
\limsup_{L\to \infty}\frac1{2\b L}\sup_{a_1,b_1}\log \cZ(a_1;\,b_1;\,   L)\leq -\t_\b(0),
  \end{equation}

 \end{lemma}
\begin{proof}
Existence and the stated properties of the surface tension are
known facts \cite[Section 4.16]{DKS}. It is also known, see  \cite[Section 4.20]{DKS}, that $\tau_\beta(\theta)$
tends to $|\cos \theta\,|+ |\sin \theta\,|$, as $\beta\to\infty$. In particular, $\t_\b(0)\to 1$, $\beta\to\infty$. Strictly speaking the proofs in \cite{DKS} are carried out for the contour ensemble associated to the 2D Ising model, which has the form \eqref{contourmod} but with slightly different potentials in the ``decoration term'' $\Phi_{L,\infty}(\g)$. However, thanks to the properties listed in Lemma \ref{lem:dks}, the same proofs actually apply to our model in \eqref{contourmod} as well.

To prove \eqref{eq:surftens2} we distinguish two cases. If $|b_1-a_1|>4L $ we use again \eqref{rev1} to obtain
$$\suptwo{a_1,b_1:}{|b_1-a_1|>4L}\cZ(a_1;\,b_1;\,   L)\leq
\sum_{\g_1: |\g_1|\ge 5L} e^{-(\b-\e(\b))|\g_1|},$$
where $\g_1$ is a contour from $(-(L+1,a_1)$ to $((L+1),b_1)$. Clearly the above sum is negligible w.r.t.\ $\exp(-2\beta
L\tau_\beta(0))$ as $L\to \infty$ for $\b$ large enough. If instead $|b_1-a_1|\leq 4L$, then the  estimate \cite[Eq.\ (4.12.3)]{DKS}
together with convexity of the surface tension allows one to conclude \eqref{eq:surftens2}.
\end{proof}
It is not hard to check that the special case $\theta=0$ coincides with the quantity in Definition \ref{deftb}.
Indeed, using the notation from Definition \ref{deftb} together with \eqref{ub3a1} (with $n=1$ and $a_1=b_1=0$),
\[
\frac{Z^\xi_{\L_L}}{Z_{\L_L}}= \frac{Z(0;0;L,L)}{Z_{\L_L}}=\textstyle\sum_{\g_1}\exp{\big(-\b|\g_1| +
 \Phi_{L,L}(\g_1)\big)}.
\]
The same arguments of Lemma \ref{lem:infvol} can be used to check that
\[
\lim_{L\to \infty} \frac 1L \log\frac{Z^\xi_{\L_L}}{Z_{\L_L}}=\lim_{L\to \infty} \frac 1L \log\cZ(0;0;L).
\]

\section{Lower  bound}\label{lb}
Here we prove the lower bound in Theorem \ref{th:main}.
We first establish a lower bound on the probability of having zero height at the boundary of a square.
\begin{lemma}\label{lem:bd}
For $\b\geq \b_0$ there exists $c_\b>0$ such that for any $L\in\bbN$:
\begin{align}\label{posp}
\bbP({\eta_{\partial\L_{L}}}=0)\geq e^{-c_\b L}.
\end{align}
\end{lemma}
\begin{proof}
Recall that $\bbP(\cdot)=\lim_{K\to\infty}\bbP_{\L_K}(\cdot)$. Expanding as in  \eqref{ub202}, we see that
\begin{eqnarray}
  \label{eq:4}
\textstyle  \bbP_{\L_K}({\eta_{\partial\L_{L}}}=0)=\frac{Z_{\L_K\setminus \partial\L_L}}{Z_{\L_K}}=\exp\left({-\sum_{V\subset
      \L_K,V\cap \partial \L_L\ne\emptyset}\varphi_0(V)}\right),
\end{eqnarray}
where $V\cap \partial \L_L\ne\emptyset$ is equivalent to $V$ not
contained in $\L_K\setminus \partial \L_L$. From the decay properties
of the potentials $\varphi_0$  stated in Lemma \ref{lem:dks}, the desired
result follows.
\end{proof}

\subsection{Proof of the lower bound in Theorem \ref{th:main}}
If we prove the lower bound for $\bbP_{\L_L}$ in \eqref{LDsos}
we also have the same lower bound for $\bbP$ by using Lemma \ref{lem:bd} and
\begin{align}\label{posp1}
\bbP( \eta_{\L_L}\geq 0) \geq   \bbP({\eta_{\partial\L_{L}}}=0)\bbP_{\L_L} ( \eta_{\L_L}\geq 0).
\end{align}
To prove the lower bound for $\bbP_{\L_L}$ we proceed by restricting the set of configurations
to an event $E$ defined as follows.
Fix $ N:=H(L)=\lfloor\tfrac1{4\b} \log L\rfloor$.

Define the nested annular regions
$\bar \cU_i:=\L_{L-3\ell_{i-1}}\setminus\L_{L-3\ell_i}$, $i=1,\dots, N$, where $\ell_0=0$ and $\ell_i=i(i+1)/2$. Notice that each $\bar \cU_i$ consists of 3 nested disjoint annuli each of width $i$. We define $\cU_i$ as the middle one, i.e.\ $\cU_i =  \L_{L-(3\ell_{i-1} -i)}\setminus\L_{L-(3\ell_i+i)}$. These sets are such that $d(\cU_i,\cU_{i+1})\geq 2i +1$, where $d(\cdot,\cdot)$ stands for the euclidean distance.

For each $i$, define the set $\cC_i$ of all contours $\g$
such that $\g\subset\cU_i$ and $\g_i$ surrounds $\L_{L-(3\ell_i+i)}$.
We consider the event $E$ that for each $i=1,\dots,N$ there exists an $i$-contour $\g_i\in\cC_i$:
\begin{align}\label{eventoE}
E = \cup_{\g_1\in\cC_1,\dots,\g_N\in\cC_N} \sC_{\g_1,1}\cap\cdots\cap\sC_{\g_N,N}.

\end{align}
 For a fixed choice of $\g_i\subset \cC_i$, $i=1,\dots, N$ we write $S_i=\L_{\g_{i-1}}\setminus \L_{\g_i}$, and
 $\D_i^+=S_i\cap \D^+_{\g_{i-1}}$, and $\D_i^-=S_i\cap \D^-_{\g_{i}}$ as in Section \ref{wedding}. We define  $Z_{S_i,\D_i^+,\D_i^-}^+$ as the partition function in $S_i$ with boundary conditions $i-1$ in $\partial S_i$, and with the following constraints: $\eta(x)\leq i-1$ for $x\in \D_i^+$, $\eta(x)\geq i-1$ for $x\in \D_i^-$ and $\eta(x)\geq 0$ for all $x\in S_i$. Then
\begin{align}\label{lb4}
\bbP_{\L_L}( \eta_{\L_L}\geq 0;\sC_{\g_{1},1}\cap\cdots\cap\sC_{\g_N,N})
=\frac{e^{-\b\sum_{i=1}^N|\g_i|}\prod_{i=1}^{N+1}Z_{S_i,\D_i^+,\D_i^-}^+}{Z_\L}.
\end{align}
Below, we shall take $n:=\lfloor\log \log L\rfloor$ and fix arbitrary contours $\g^*_1\in\cC_1,\dots,\g^*_n\in\cC_n$, and sum over  the remaining contours $\g_i$, $i=n+1,\dots,N$
 \begin{lemma}\label{lemlb}
 Fix $\b\geq \b_0$ and fix  $\g^*_1\in\cC_1,\dots,\g^*_n\in\cC_n$, where $n=\lfloor\log \log L\rfloor$. Then
  \begin{align}\label{lb1}
 \bbP_{\L_L}( \eta_{\L_L}\geq 0; E) \geq\; \tfrac12\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{\g_{n+1}\in\cC_{n+1},\dots,\g_N\in\cC_N}\!\!\!\!\!\!\!\!\!\!\!\!
 \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}

 ).
 \end{align}
 \end{lemma}
\begin{proof}
Let $F_i$ denote the event that there is more than
one $i$-contour in $\cC_i$.
Then
\begin{align}\label{lb2}
\bbP_{\L_L}( \eta_{\L_L}\geq 0; E)  \geq
\!\!\!\!\!\!\!\!\!\sum_{\g_{n+1}\in\cC_{n+1},\dots,\g_N\in\cC_N}\!\!\!\!\!\!\!
 \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}

 \,;\,\cap_{i=n+1}^NF_i^c).
\end{align}
Thus, it suffices to show that for any fixed choice of $\g^*_k\in\cC_k$, $k=1,\dots,n$ and $\g_j\subset \cC_j$, $j=n+1,\dots, N$:
\begin{align}\label{lb3}
& \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}\,;\,\cup_{i=n+1}^NF_i)\nonumber
\\ & \qquad \qquad\qquad \qquad \leq\tfrac12\, \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}).
\end{align}

Suppose the $j$-contour $\g_j\in\cC_j$ is given for each  $j=n+1,\dots,N$. If $F_i$
occurs then there must be a $i$-contour $\g\in\cC_i$, $\g\neq\g_i$, such that either $\g\subset S_{i+1}$ or
$\g\subset S_{i}$. In particular, if $\cup_{i=n+1}^NF_i$ occurs, then,  for some $i\in[ n+1,N+1]$,
there exists either an $(i-1)$-contour or an $i$-contour $\g$
inside $S_i$ and surrounding $\L_{L-(3\ell_i+i)}$. Let $\pi_{S_i,\D_i^+,\D_i^-}^+$ denote the probability measure corresponding to the
partition function $Z_{S_i,\D_i^+,\D_i^-}^+$. From \cite[Proposition 2.7]{CLMST2} one has
that for any fixed contour $\g$ inside $S_i$, for any $h\in\bbN$:
\begin{align}\label{lb5}
\pi_{S_i,\D_i^+,\D_i^-}^+(\sC_{\g,h}) \leq \exp{\big(-\b|\g| + Ce^{-4\b h}|S_i| + Ce^{-4\b h}|\g|\log|\g|
\big)}.
\end{align}
Here and below, by $C$ we mean a positive constant that does not depend on $\b$ and $L$, whose value may change at each occurence.  Since $|S_i| \leq CL i\leq  L\log L$,
and  $\log|\g|\leq 2\log L$, taking either $h=i$ or $h=i-1$, with $i\geq n+1$ one has that
$e^{-4\b h}|S_i|\leq L(\log L)^{1-4\b}$ and $e^{-4\b h}\log|\g|\leq 2(\log L)^{1-4\b}$, and therefore
 \begin{align}\label{lb505}
\pi_{S_i,\D_i^+,\D_i^-}^+(\sC_{\g,h}) \leq
\exp{\big(-(\b-1)|\g| + L
\big)},
\end{align}
as soon as $\b$ and $L$ are large enough. If $\g$ is required to surround $\L_{L-3\ell_i+i}$, then necessarily $|\g|\geq 2L$. Let $p_i$ denote the $\pi_{S_i,\D_i^+,\D_i^-}^+$-probability that there exists either an $(i-1)$-contour or an $i$-contour $\g$
inside $S_i$ and surrounding $\L_{L-3\ell_i+i}$.
Summing over $\g\subset S_i$ with $|\g|\geq 2L$ in \eqref{lb505}, one finds that for $\b$ large enough,
$p_i\leq e^{-L}$. From \eqref{lb4}, using a union bound and the fact that $Ne^{-L}\leq 1/2$, it follows that
\begin{align}\label{lb6}
&  \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}\,;\,\cup_{i=n+1}^NF_i)\nonumber \\ & \qquad \qquad \leq \sum_{i=n+1}^N p_i \, \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j})\nonumber\\
 &\qquad \qquad \qquad \qquad  \leq \tfrac12\, \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}).
 \end{align}
\end{proof}
Thanks to Lemma \ref{lemlb} the lower bound in Theorem \ref{th:main} follows if we prove
that
\begin{align}\label{lb7}
\!\!\!\!\!\!\!\! \sum_{\g_{n+1}\in\cC_{n+1},\dots,\g_N\in\cC_N}\!\!\!\!\!\!\!\!\!\!\!\!
&  \bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j})  \nonumber \\
  & \qquad\qquad \quad \geq \exp{\big(-8\b\t_\b(0)NL(1+o(1))\big)},
 \end{align}
 for any fixed choice of $\g^*_k\in\cC_k$, $k=1,\dots,n$, with $n=\lfloor\log \log L\rfloor$.
 To prove \eqref{lb7} we start by observing that by the FKG inequality
 one has
  \begin{align}\label{lb50}
\frac{Z_{S_i,\D_i^+,\D_i^-}^+}{Z_{S_i,\D_i^+,\D_i^-}} =
\pi_{S_i,\D_i^+,\D_i^-}(\eta(x)\geq 0,\,\forall x\in S_{i})\geq
\prod_{x\in S_{i}}\pi_{S_i,\D_i^+,\D_i^-}(\eta(x)\geq 0)\,,
\end{align}
where $Z_{S_i,\D_i^+,\D_i^-}^+$ is defined above \eqref{lb4}, $Z_{S_i,\D_i^+,\D_i^-}$ is as in Section \ref{cl_exp}, and $\pi_{S_i,\D_i^+,\D_i^-}$ denotes the probability measure associated to the partition function $Z_{S_i,\D_i^+,\D_i^-}$. From \cite[Proposition 3.9]{CLMST} one has that $\pi_{S_i,\D_i^+,\D_i^-}(\eta(x)\geq 0)\geq 1-Ce^{-4\b (i-1)}$ for any $x\in S_i$. Using $1-x\geq e^{-2x}$ for $0\leq x\leq 1/2$, one has
   \begin{align}\label{lb51}
\frac{Z_{S_i,\D_i^+,\D_i^-}^+}{Z_{S_i,\D_i^+,\D_i^-}}
\geq \exp{\big(-2C|S_i|e^{-4\b (i-1)}\big)}\,.
\end{align}
Therefore, in \eqref{lb4} we can estimate
\begin{align}\label{lb40}
&\bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j})\nonumber \\ & \qquad
\geq \textstyle \exp{\big(-\b\sum_{i=1}^n|\g^*_i|-\b\sum_{i=n+1}^N|\g_i|-2C\sum_{i=1}^{N+1}|S_i|e^{-4\b (i-1)}\big)}
\frac{\prod_{i=1}^{N+1}Z_{S_i,\D_i^+,\D_i^-}}{Z_\L}.
\end{align}
Expanding as in \eqref{ub2a1} one obtains
\begin{align}\label{lb52}
\frac
{\prod_{i=1}^{N+1}Z_{S_i,\D_i^+,\D_i^-}}{Z_\L}
=\textstyle \exp{\big(
 \Psi_\L(\g^*_1,\dots,\g^*_n,\g_{n+1},\dots,\g_N)\big)},
\end{align}
where $\Psi$ is given in \eqref{ub211}.
 Estimating $|S_i|\leq Ci L$ one finds
 \begin{align}\label{lb53}
\sum_{i=1}^{N} |S_i|e^{-4\b (i-1)}\leq CL.
 \end{align}
 On the other hand, the term $|S_{N+1}|e^{-4\b N} = |\L_{\g_N}|e^{-4\b H(L)}$ satisfies
 \begin{align}\label{lb533}
|S_{N+1}|e^{-4\b N}\leq L^2 e^{-4\b H(L)} \leq
CL,
 \end{align}
 where we use $e^{-4\b H(L)}\leq C/L$. Note that it is at this point of the argument that it is crucial to have $N$ as large as $H(L)$.
From \eqref{lb53}-\eqref{lb533} one has $\sum_{i=1}^{N+1} |S_i|e^{-4\b (i-1)}\leq CL$.
From this bound and \eqref{lb52} we obtain
\begin{align}\label{lb400}
&\bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{k=1}^n\sC_{\g^*_{k},k}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j})\nonumber \\ &
\geq \textstyle \exp{\big(-\b\sum_{i=1}^n|\g^*_i|-\b\sum_{i=n+1}^N|\g_i| + \Psi_\L(\g^*_1,\dots,\g^*_n,\g_{n+1},\dots,\g_N) -  CL \big)}.
\end{align}
  Next, we want to show that the interactions among the contours are
  negligible in our setting.
 Let $\psi_\L(\g)$ denote the potential associated to a single contour $\g$ as defined in \eqref{ub1c}.
 \begin{lemma}\label{lemlb1} Take $\b\geq \b_0$. Uniformly in the choice of $\g_1\in \cC_1,\dots,\g_N\in\cC_N$ one has
 \begin{align}\label{lb54}
\textstyle\left|\Psi_\L(\g_1,\dots,\g_N)-\sum_{i=1}^N\psi_\L(\g_i)\right|\leq \sum_{i=1}^{N} |\g_i|e^{-\b i/2}.
 \end{align}
 \end{lemma}
 \begin{proof}
 Notice that any $V\subset \L$ such that $d(V,\g_i)\leq 1$ and $d(V,\g_{i+1})\leq 1$
 must have $d(V)\geq 2i$. Thus the sum of the potentials associated to $V$'s that have $d(V,\g_i)\leq 1$ and
 are such that $d(V,\g_j)\leq 1$ for some $j\neq i$ contributes at most $|\g_i|e^{-\b i/2}$
 if $\b$ is large enough.
 \end{proof}

From \eqref{lb400} and Lemma \ref{lemlb1}
one has
\begin{align}\label{lb55}
&\bbP_{\L_L}( \eta_{\L_L}\geq 0\,;\,\cap_{i=1}^n\sC_{\g^*_{i},i}\,;\,\cap_{j=n+1}^N\sC_{\g_{j},j}

)\nonumber \\ &
\geq \textstyle \exp{\big(-2\b\sum_{i=1}^n|\g^*_i|-\b\sum_{i=n+1}^N|\g_i|(1+e^{-\b i/2}) +\sum_{i=1}^N\psi_\L(\g_i) -CL\big)}
.
\end{align}
For $i=1,\dots,n$, we can use the rough estimates $|\g_i|\leq CL n \leq  C L \log \log L$
and $|\psi_\L(\g_i)| \leq C|\g_i|$ (cf. \eqref{rev1}) to obtain
\begin{align}\label{lb56}
\textstyle \exp{\big(-2\b\sum_{i=1}^n|\g_i|+\sum_{i=1}^n\psi_\L(\g_i)\big)}
\geq \exp (-o(L\log L)).
\end{align}
For $n<i\leq N$ we need the following statement.
\begin{lemma}\label{lemlb2} Uniformly over $i$ such that $n<i\leq N$, one has
 \begin{align}\label{lb57}
\sum_{\g\in\cC_i} \exp{\big(-\b|\g|(1+e^{-\b i/2}) +\psi_\L(\g)\big)}
\geq \exp{\big(-8\b\t_\b(0) L(1+o(1))

\big)}  .
\end{align}
 \end{lemma}
We first conclude the proof of the lower bound in Theorem \ref{th:main}, assuming the estimate of Lemma \ref{lemlb2}. From Lemma \ref{lemlb} and \eqref{lb55}-\eqref{lb56}
we have
\begin{align}\label{lb58}
\bbP_{\L_L}( \eta_{\L_L}\geq0)&\geq \bbP_{\L_L}( \eta_{\L_L}\geq 0; E)
\geq \exp (-o(L\log L))\times \nonumber\\
&\times
\textstyle\sum_{\g_{n+1}\in\cC_{n+1},\dots,\g_N\in\cC_N}\exp{\big(-\b\sum_{i=n+1}^N|\g_i|(1+e^{-\b i/2}) +\sum_{i=1}^N\psi_\L(\g_i)\big).
}\end{align}
 From Lemma \ref{lemlb2} and using $NL=1/(4\b)L\log L+O(L)$ one has
 \begin{align}\label{lb59}
\bbP_{\L_L}( \eta_{\L_L}\geq0)
\geq\exp{\big(-8\b\t_\b(0) NL(1+o(1))
\big)}  .
\end{align}
This concludes the proof.

 \begin{proof}[Proof of Lemma \ref{lemlb2}]
 First observe that $\g\in\cC_i$ implies $|\g|\leq |S_i|\leq L\log L$ and therefore
 for $i\geq \log\log L$ and $\b\geq \b_0$ one has
 $$|\g|e^{-\b i/2} = o(L).$$
Next, observe that we may safely replace $\psi_\L(\g)$ in \eqref{lb57} by the quantity $\psi_\infty(\g)$ (see the end of Section \ref{wedding}). Indeed, any connected set $V$ that touches both $\cU_i$ and $\partial \L$ must have $d(V)\geq \tfrac12(\log \log L)^2$.
Thus, we have to show that
 \begin{align}\label{lb60}
\sum_{\g\in\cC_i} \exp{\big(-\b|\g| +\psi_\infty(\g)\big)}
\geq \exp{\big(-8\b\t_\b(0) L(1+o(1))

\big)}  .
\end{align}
To prove \eqref{lb60} we fix $i$ and partition the set $\cU_i$ into rectangles $R_j$, $j=1,\dots,m$,
with height $i$ and basis $i^{2-\e}$, so that  there are $m\sim
8Li^{-2+\e}$ such rectangles, see Figure \ref{fig:05}. For simplicity, let us assume that the partitioning is exact so that $\cU_i$ is the union of the $R_j$'s plus four squares at the corners as in Figure \ref{fig:05}. The modifications in the general case are straightforward.
\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.47]{annulus}
\put(46,18) {$x_j$}
\put(101.3,18) {$y_j$}
\put(70,16) {$\hat \g_j$}

\end{overpic}

        \caption{
        The partition of $\cU_i$ into rectangles $R_j$, $j=1,\dots, m$ (left). A single path $\hat \g_j:x_j\to y_j$ inside the rectangle $R_j$ (right).
          }\label{fig:05}
\end{figure}

We fix for every rectangle $R_j$ the points $x_j$ and $y_j$ that are
the midpoints of the two shorter side. Consider an open contour
$\hat\g_j$ connecting $x_j$ to $y_j$ which is entirely contained in
$R_j$ (see Figure \ref{fig:05}).  For technical reasons it is convenient to consider a closed path $\hat \g$ that agrees with $\hat\g_j$ on $R_j$. The latter is defined as follows. Let $\hat \g $ be the closed contour
contained in $\cU_i$ which coincides with $\hat\g_j$ inside $R_j$,
it is given by straight segments in all other rectangles $R_k$, $k\neq j$, and by a straight right angle shape at each of the four corner squares. Then
we define $\psi_\infty(\hat\g_j)$ as $\psi_\infty(\hat \g)$ (see text after \eqref{rev1}) but with the restriction to those sets $V$ which have
distance from $\hat\g_j$ at most $1$.

It follows from  \cite[Sections 4.12 and 4.15]{DKS} that for a fixed index $j$ one has, for $i$ large:

  \begin{align}\label{lb61}
\sum_{ \hat \g_j\,:\;x_j\to y_j,\;\hat \g_j\subset R_j}
\exp{\big(-\b| \hat \g_j| +\psi_\infty( \hat\g_j)\big)}
\geq \exp{\big(-\b\t_\b(0) i^{2-\e}(1+o(1))

\big)}.
\end{align}
The point is that the height $i$ of the rectangles $R_j$ is much
larger than the typical vertical fluctuation $i^{1-\gep/2}$ of paths
$\hat\gamma_j$, so the restriction to be in $R_j$ is not modifying the
partition function significantly.

Suppose now that $\g\in\cC_i$ is a contour
 passing through all the points $x_j,y_j
$ that can be written as the composition of $\hat\g_1,\dots,\hat\g_m$ where $ \hat\g_j$ is as in the sum above, and assume that it has some prescribed shape at the four corners of $\cU_i$, e.g.\ a right angle form as in
Figure \ref{fig:05}.
Then it is immediate to check that
$|\g|\leq \sum_{j=1}^m|\hat\g_j| + O(i) $, and
$$\psi_\infty(\g) - \sum_{i=1}^m\psi_\infty(\hat \g_j) = O(i m)\,.$$
The latter estimate holds thanks to the decay properties of the
potentials, so that the mutual interaction between  $\hat\gamma_j$ and
$\hat\gamma_{j-1}$ is $O(i)$ uniformly in $j=1,\dots, m$.
Thus, by restricting the sum in \eqref{lb60} to contours as in \eqref{lb61}
one obtains
\begin{align}\label{lb62}
\sum_{\g\in\cC_i} \exp{\big(-\b|\g| +\psi_\infty(\g)\big)}
\geq \exp{\big(-\b\t_\b(0) m \,i^{2-\e}(1+o(1))

\big)}  .
\end{align}
Since $m\sim 8Li^{-2+\e}$, the desired estimate follows.
\end{proof}

\section{A monotonicity property of the SOS model}\label{monostate}
Recall the staircase ensemble defined in Section \ref{stair} with partition function
$$\cZ(a_1,\dots,a_n; \,b_1,\dots,\, b_n; L),$$
as defined in Lemma \ref{lem:infvol}.
In this section we establish the following important monotonicity property.
\begin{theorem}
\label{th:m1}
There exists $\b_0>0$ such that, for any $\b>\b_0$ and any $L\in \bbN$
  \begin{gather}
    \label{eq:2}
\cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\, L)\leq \prod_{i=1}^n \cZ(a_i;\, b_i;\,
  L).
  \end{gather}
\end{theorem}
The above estimate allows one to control the partition function of $n$ interacting open contours by means of the partition functions of $n$ non-interacting open contours. In particular, Theorem \ref{th:m1} and Lemma \ref{lem:surftens} yield the following corollary.

\begin{corollary}\label{coro_surftens}
Fix $n\in\bbN$, and suppose that as $L\to\infty$ one has $(b_i-a_i)/L\to \l_i \in\bbR$, $i=1,\dots,n$. Then
\begin{align}\label{coros1}
\limsup_{L\to\infty}\frac1{2L}\log \cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\, L) \leq -\b\sum_{i=1}^n\frac{\t_\b(\theta_i)}{\cos(\theta_i)}
\end{align}
where $\theta_i=\tan^{-1}(\l_i)$.

\end{corollary}

The proof of Theorem \ref{th:m1} is based on the following key lemma.

\begin{lemma}
\label{lem:1}
Given $\{a_i,b_i\}_{i=1}^n$,  let $\{a'_i,b'_i\}_{i=1}^n$
  be defined by
\[
a'_i=a_i,\
b'_i=b_i\,,\quad i=1,\dots,n-1;\quad a'_n=a_n+1,\ b'_n=b_n+1.
\]
Then
 \[
\cZ(a_1,\dots,a_n;\ b_1,\dots,b_n;\
  L)\leq \cZ(a'_1,\dots,\,\,a'_n;\ b'_1,\dots,\, b'_n;\
  L).
\]
\end{lemma}
\begin{proof}[Proof of  Lemma \ref{lem:1}]
Set $\L:= \L_{L,M}$ for some large fixed $M>\max\{a_n,b_n,-a_1,-b_1\}$. Let
$\t,\t'$ be the SOS boundary conditions associated to
$\{a_i,b_i\}_{i=1}^n$ and $\{a'_i,b'_i\}_{i=1}^n$ according to
\eqref{stairtau}.  Given $s\in [0,1]$ consider the auxiliary boundary condition
$\tau_s: \partial \L\mapsto \bbR$ defined by
\begin{equation*}
 \tau_s(x_1,x_2)=
 \begin{cases} n-1+s & \text{ if $(x_1,x_2)=(-L-1,a_n)$ or $(x_1,x_2)=(L+1,b_n)$}; \\
   \tau'(x_1,x_2)
   &\text{ otherwise}.
 \end{cases}
\end{equation*}
Next, we  consider the partition function $Z_\L^{\tau_s}$ associated to $\t_s$
(strictly speaking we have only defined the model for integer valued boundary condition, but it is straightforward to extend it to the real valued case).
Notice that $\t_s = s\t+(1-s)\t'$. We shall see that
$Z_\L^{\tau_s}$ is differentiable w.r.t.\ $s\in[0,1]$ so that
\begin{equation}\label{diffz}
Z_\L^\t-Z_\L^{\t'}= \int_0^1 ds \, \frac{d}{ds}Z_\L^{\tau_s}.
\end{equation}
In order to compute the above derivative we proceed as follows.
Define the points
$z=(-(L+1),a_n),w=(-L,a_n)$ and  $z'=(L+1,b_n),\
w'=(L,b_n)$, so that $w$ (resp.\ $w'$) is the nearest neighbor of $z$ (resp.\ $z'$) in $\L$, see Figure \ref{fig:001}.
\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.3]{fig_top2}
\put(94.5,77) {$z'$}
\put(2,57){$z$}
\put(8,57){$w$}
\put(86.5,77) {$w'$}

\put(2,2.3){$0$}
\put(95,30){$0$}
\put(2,34.5){$1$}
\put(95,54.5){$1$}
\put(2,71){$2$}
\put(95,87){$2$}
\end{overpic}
        \caption{A sketch of the staircase boundary condition with $n=2$ steps as seen from
               above, with two open contours and the pairs of vertices appearing in the proof of  Lemma \ref{lem:1}:
          $z=(-L-1,a_2), w=(-L,a_2)$, $z'=(L+1,b_2), w'=(L+1,b_2)$.}\label{fig:001}
\end{figure}

Let $\cB_\L^*=\cB_\L\setminus\{wz,w'z'\}$ denote all bonds with at least one vertex in $\L$

with the exception of the two bonds $wz$ and $w'z'$.
Define the energy function $\cH_\L^{\t,*}(\eta),\ \eta\in \O_\L^{\t_s}$ by
\[
\cH_\L^{\t,*}(\eta)= \sum_{xy\in\cB_\L^*}|\eta(x)-\eta(y)| + \phi(\eta(w)) + \phi(\eta(w')),
\]
where
\[
\phi(h)= (h-n){\mathds 1}_{\{h\geq n\}} + (n-1-h){\mathds 1}_{\{h\leq n-1\}},\quad
h\in \bbZ.
\]
Since the bonds $wz$ and $w'z'$ are not included in the above sum, we see that $\cH_\L^{\t,*}(\eta)$ does not depend on the parameter $s$. Let also
\[
F_{s,n}(\eta):= \exp\left(-\b\bigl[(1-s)({\mathds
         1}_{\{\eta(w)\geq n\}}+{\mathds
         1}_{\{\eta(w')\geq n\}})+s({\mathds
         1}_{\{\eta(w)\leq n-1\}}+{\mathds
         1}_{\{\eta(w')\leq n-1\}})\bigr]\right).
\]
Define the partition function $\Xi^{\t,*}_\L= \sum_{\eta\in\O_\L^{\t_s}}\exp(-\b \cH_\L^{\t,*}(\eta))$, and the Gibbs measure $$\pi_\L^{\t,*}(\eta)=(\Xi^{\t,*}_\L)^{-1} \exp(-\b \cH_\L^{\t,*}(\eta)),$$ $\eta\in\O_\L^{\t_s}$.
It is not hard to check that
\[
     Z_\L^{\tau_s}= \Xi^{\t,*}_\L \, \pi_\L^{\t,*}\left(F_{s,n}\right).
\]

Using the above expression for $Z_\L^{\tau_s}$ we get
\begin{align*}
\frac{d}{ds} Z_\L^{\tau_s}=  \Xi^{\t,*}_\L \,\pi_\L^{\t,*}\left(\frac{d}{ds} F_{s,n}\right)=
\b\, \Xi^{\t,*}_\L \,\pi_\L^{\t,*}\left(G_s\right),

\end{align*}
where, for any $s\in [0,1]$, we define
\[
G_{s,n}(\eta):= F_{s,n}(\eta)\left({\mathds
         1}_{\{\eta(w)\geq n\}}+{\mathds
         1}_{\{\eta(w')\geq n\}} - {\mathds
         1}_{\{\eta(w)\leq n-1\}}-{\mathds
         1}_{\{\eta(w')\leq n-1\}}\right).
\]
The function $G_{s,n}$ takes values in $\{-2e^{-2\b s},0,2e^{-2\b(1-s)}\}$ and is easily seen to be  increasing in the configuration $\eta$.
Therefore, if we raise to height $n-1$ the value of $\t$
on those boundary vertices where it was at most $n-1$ and we denote by $\hat \t$ the resulting boundary
condition, from the FKG inequality we get that
\begin{gather*}
\pi_\L^{\t,*}\left(G_{s,n}\right)  \leq
\pi_\L^{\hat\t,*}\left(G_{s,n} \right).
\end{gather*}
The validity of the FKG inequality follows from lattice condition \eqref{holley} for the measure
$\pi_\L^{\t,*}$, which can be verified directly.
The boundary height $\hat\t$ has now a single step from level $n-1$ to level $n$. Using vertical translation invariance we can now safely replace the
height of $\hat \t$ by $0,1$ instead of $n-1,n$. Finally,
since $G_{s,1}$ is a bounded local function, we can take the limit
$M\to \infty$ in \eqref{diffz} and get that
\begin{gather*}
  \cZ(a_1,\dots,a_n;\ b_1,\dots,b_n;\
  L)- \cZ(a_1,\dots,a_n+1;\ b_1,\dots,b_n+1;\ L)\\
\leq \b\left(\lim_{M\to \infty}\frac{\Xi^{\t,*}_\L}{Z_\L}\right) \int_0^1
ds\, \pi^{\hat \t,*}_\infty\left(G_{s,1} \right),
\end{gather*}
where $\pi^{\hat \t,*}_\infty(\cdot)$ denotes the weak limit as $M\to\infty$ of $\pi_\L^{\hat\t,*}$, that is the Gibbs measure on $\L_{L,\infty}=[-L,L]\times \bbZ$ with boundary
condition at height 1 at the vertices  $x=(x_1,x_2)$ with
either $x_1=-(L+1)$ and
$x_2\geq a_n+1$ or $x_1=L+1$ and
$x_2\geq b_n+1$; the boundary height is unspecified at the vertices
$z,z'$
(this simply means that the terms corresponding to bonds $wz$ and $w'z'$ do not appear in the interaction) and otherwise it is equal to zero.
The existence of the limits mentioned above can be proved again from the cluster expansion representation as in Lemma \ref{lem:infvol}.
By symmetry one has that
\[
\pi_\infty^{\hat \tau,*}(\eta(w)\geq 1;\, \eta(w')\geq 1)=
\pi_\infty^{\hat \tau,*}(\eta(w)\leq 0;\,  \eta(w')\leq 0),
\]
so that
\[
\pi_\infty^{\hat \tau,*}\left(G_{s,1} \right)=-\pi_\infty^{\hat \tau,*}\left(G_{1-s,1} \right)  \quad
\text{ and }\quad \int_0^1ds\, \pi_\infty^{\hat \tau,*}\left(G_{s,1} \right)=0.
\]
In conclusion
\[
  \cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\,
  L)\leq \cZ(a'_1,\dots,a'_n;\, b'_1,\dots,b'_n;\, L)
\]
and the lemma is proved.\end{proof}

We can now complete the proof of Theorem \ref{th:m1}. By iterating Lemma
\ref{lem:1} arbitrarily many times, we have that
\[
   \cZ(a_1,\dots,a_n;\, b_1,\dots,b_n;\, L)\leq \lim_{k\to \infty}
  \cZ(a_1,\dots,a_{n-1},a_n+k;\, b_1,\dots,b_{n-1},b_n+k;\,
  L).
\]
On the other hand, using the explicit representation \eqref{ub4a1}

together
with the rough bound  \eqref{rev4} to control the large deviations of the
$n$-th contour $\g_n$, we have
that
\begin{gather*}
\lim_{k\to \infty}
  \cZ(a_1,\dots,a_{n-1},a_n+k;\, b_1,\dots,b_{n-1},b_n+k;\,
  L)\\=
  \cZ(a_1,\dots,a_{n-1};\, b_1,\dots,b_{n-1};\, L)\cZ(a_n;\ b_n;\ L).
\end{gather*}
In conclusion, we have factorized out the contribution of the $n$-th
contour. By repeating the above reasoning for $(a_{n-1},b_{n-1}),\,(a_{n-2},b_{n-2})\dots
,(a_{2},b_{2})$ we finally get \eqref{eq:2}.

\section{Upper bound}\label{ub}
If we prove the upper bound for $\bbP$ in \eqref{LDsos}, then we can obtain the upper bound for
$\bbP_{\L_L}$ by using \eqref{posp1} and Lemma \ref{lem:bd}. From now on  we concentrate on proving the upper bound for $\bbP$.

For any event $A$, note that
\begin{align}\label{ub1}
\bbP( \eta_{\L_L}\geq 0) \leq
\frac{\bbP(A)}{\bbP(A\tc \eta_{\L_L}\geq 0)}.
\end{align}
Indeed, \eqref{ub1} is obtained by multipling by $\bbP( \eta_{\L_L}\geq 0)$ both sides of the obvious inequality $1\leq \bbP(A)/
\bbP(A,\eta_{\L_L}\geq 0)$.

For any $\d>0$ and $K>0$, define $A=A(\d,K)$, as the event that there exists a lattice circuit $\cC$ surrounding $\L':=\L_{(1-\d)L}$
such that $\eta(x)\geq H(L) - K$, for all $x\in\cC$, where as usual $H(L)=\lfloor\tfrac1{4\b} \log L\rfloor$.
\begin{proposition}\label{propoad1}
For any $\d>0$, there exists a constant $K>0$ such that
\begin{align}\label{adub2}
\lim_{L\to\infty}\bbP(A(\d,K)\tc \eta_{\L_L}\geq 0)= 1.
\end{align}
\end{proposition}
\begin{proof}
Let $\partial_* \L_L$ denote the internal boundary of $\L_L$.
Observe that $A(\d,K)$ is monotone increasing so that by the FKG inequality $$\bbP(A(\d,K)\tc \eta_{\L_L}\geq 0)\geq \bbP(A(\d,K)\tc \eta_{\L_L}\geq 0, \,{\eta_{\partial_*\L_{L}}}=0).$$
Therefore, the proposition follows once we know that for some $K=K(\d)$ one has
\begin{align}\label{adub02}
\lim_{L\to\infty}\bbP(A(\d,K)\tc \eta_{\L_L}\geq 0, \,{\eta_{\partial_*\L_{L}}}=0)= 1.
\end{align}
Under the conditioning $\eta_{\L_L}\geq 0,
\,{\eta_{\partial_*\L_{L}}}=0$, one has an SOS interface in $\L_{L-1}$
with a wall at height zero and zero boundary conditions. The result of \cite[Theorem 2]{CLMST2}
implies that with probability converging to $1$, within $\L_{L-1}$, there exists an $h$-contour surrounding $\L'=\L_{(1-\d)L}$, for all $h\leq H(L) -K$ as soon as $K$ is a sufficiently large constant depending on $\d$.
This implies \eqref{adub02}.
\end{proof}

It follows that to prove the upper bound in \eqref{LDsos} it is sufficient to establish:
\begin{proposition}\label{propoad}
For any $\d>0$, for any $K>0$, one has
\begin{align}\label{ub2}
\limsup_{L\to\infty}\frac1{2L\log L}\log \bbP( A(\d,K)) \leq  -\t_\b(0)(1-\d).
\end{align}
\end{proposition}

\subsection{Proof of Proposition \ref{propoad}} The first observation
is that we may impose zero boundary conditions outside a very large
set, e.g.\ $\L_M$ with $M\gg L^2$, and therefore we may consider
$\wt\bbP:=\bbP_{\L_M}$ instead of $\bbP$ in \eqref{ub2}.
The reason is that the probability that there is a contour surrounding
$\L'$ and not contained in, say, $\L_{L^2}$ is a negligible
$O(\exp(-L^2))$, as one can check easily using a rough estimate as in \eqref{rev1}. Then, $A(\d,K)$ can be
considered as a local event (localized in $\L_{L^2}$) and by
definition of thermodynamic limit one can approximate arbitrarily well
$\bbP( A(\d,K)) $ by $\wt\bbP( A(\d,K)) $, if $M$ is sufficiently large.

The event $A(\d,K)$ implies that for each $h=1,\dots,N:=H(L) - K$ there exists (at least) one  $h$-contour  surrounding $\L'$.
Therefore, there must exist $\L_M\supset \g_1\supset\cdots \supset\g_N\supset \L'$ such that $\g_h$ is an $h$-contour:
\begin{align}\label{ub20}
\wt \bbP( A(\d,K)) \leq \sum_{  \g_1\supset\cdots \supset\g_N\supset \L'} \wt \bbP\big( \cap_{i=1}^N\sC_{\g_i,i}\big).
\end{align}
Here we use the notation $\L_M\supset \g_1\supset\cdots \supset\g_N\supset \L'$ when the contours satisfy $\L_M\supset \L_{\g_1}\supset\cdots \supset\L_{\g_N}\supset \L'$.

For a fixed choice of $\g_1\supset\cdots \supset\g_N$ the above probability is computed in \eqref{ub2a1}:
\begin{align}\label{ub21}
\wt \bbP\big( \cap_{i=1}^N\sC_{\g_i,i}\big)=
\textstyle \exp{\big(-\b\sum_{i=1}^N|\g_i| +
 \Psi_{\L_M}(\g_1,\dots,\g_N)\big)}.
\end{align}

To deal with the  summation in \eqref{ub20}  we consider a decomposition of each contour into four ``irreducible" pieces, which will be responsible for the main contributions, plus some negligible corner terms.

\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.37]{redconts2}
\put(22.2,84.3) {$x^t$}
\put(77.3,82.2) {$y^t$}
\put(50,89.8) {$\g^t$}
\put(16,51) {$\g^\ell$}
\put(15,69.2) {$y^\ell$}
\put(17.5,23) {$x^\ell$}
\put(22.5,5) {$y^b$}
\put(72.9,1.6) {$x^b$}
\put(50,6) {$\g^b$}
\put(93,22) {$y^r$}
\put(82,70) {$x^r$}
\put(102,50) {$\g^r$}
\end{overpic}

        \caption{
        Example of a contour $\g$ surrounding the square $\L'$. The irreducible components of $\g$ are the thicker paths.
          }\label{fig:2}
\end{figure}

Let $\cS_v$ and $\cS_h$ denote, respectively, the vertical and horizontal infinite strips obtained by prolonging the sides of the square $\L'$:
\begin{align*}
\cS_v&=\{x=(x_1,x_2)\in\bbZ^2:\;|x_1|\leq (1-\d)L\}\,,\;\\
\cS_h&=\{x=(x_1,x_2)\in\bbZ^2:\;|x_2|\leq (1-\d)L\}.
\end{align*}
Let $\cS_v^t$, resp. $\cS_v^b$, denote the top, resp.\ bottom part of $\cS_v$, i.e.\ the part that lies above, resp. below, the square $\L'$. Similarly, let $\cS_h^\ell$, resp.\ $\cS_h^r$, denote the portion of $\cS_h$ to the left, resp. to the right, of the square $\L'$.

We now define the irreducible components of a fixed  contour $\g$ containing $\L'$.
Consider the portion of $\g$ that intersects $ \cS_v^t$.
This must contain at least one {\em crossing}, defined as an open contour connecting the opposite vertical sides of $\cS_v^t$ that is fully contained in the interior of $\cS_v^t$. Let $\g^t$ denote the most internal crossing, i.e.\ the one that lies closest to the square $\L'$. We repeat the same construction in the strips $\cS_h^\ell, \cS_v^b$ and $S_h^r$, to define $\g^\ell, \g^b$ and $\g^r$ as the most internal crossings. We say that $\g^u$, $u\in\{t,\ell,b,r\}$,

form the {\em irreducible components} of the contour $\g$.
We call $x^u,y^u$ the endpoints of $\g^u$, with $x^u$ coming after $y^u$ if $\g^u$ is given a counter clockwise orientation.

See Figure \ref{fig:2}.
It is easy to convince oneself that any contour
containing the square $\L'$, such that
its irreducible components coincide with the given $\g^t,\g^\ell, \g^b,\g^r$, must have the following property:
If we travel along $\g^t$
in the direction $y^t \to x^t$, and then follow the contour,
the irreducible components we meet are, in order:  $\g^\ell$
in the direction $y^\ell \to x^\ell$,  then $\g^b$ in the direction $y^b \to x^b$, then $\g^r$ in the direction $y^r \to x^r$, and finally again $\g^t$
in the direction $y^t \to x^t$.

Thus we can write any $\g$ with given irreducible components $\g^t,\g^\ell, \g^b,\g^r$ as the composition
\begin{align}\label{ub22}
&\g= \g^t\circ \h^{t,\ell}\circ \g^\ell\circ \h^{\ell,b}\circ\g^b\circ \h^{b,r}\circ\g^r\circ \h^{r,t},
\end{align}
where $\h^{u,v}$ denotes a path connecting $x^u$ and $y^v$ for $u,v\in\{t,\ell,b,r\}$.

Let
$\g_1,\dotsm\g_N$ denote a collection of nested contours as in \eqref{ub20}. We write $\g_i^u$ for the corresponding irreducible components, and $\eta_i^{u,v}$ for the remaining components.
Clearly, by applying the decomposition \eqref{ub22} for each $i$, one has
\begin{align}\label{ub23}
&
\textstyle|\g_i|=  |\h_i^{t,\ell}|+|\h_i^{\ell,b}|+|\h_i^{b,r}|+| \h_i^{r,t}| + \sum_u|\g_i^u|,

\end{align}
where the sum ranges over $u\in\{t,\ell,b,r\}$.

Next, we want to decouple the four irreducible pieces, by writing
$\Psi_{\L_M}(\g_1,\dots,\g_N)$ as the sum of a main term $\sum_u\Psi_u(\g^u_1,\dots,\g^u_N)$ and a correction term associated to the corner pieces $\eta_i$ and to the interactions between distinct irreducible regions. To this end it will be convenient to enlarge  the strips $\cS_v,\cS_h$ by an amount of order $(\log L)^2$. This will ensure that the expression \eqref{ub21} factorizes (up to lower order terms) into  the product of four distinct pieces which, see  Lemma \ref{prozs} below, can each be reinterpreted as probabilities from the SOS staircase ensemble defined in Section \ref{stair}.
 To define the potential  $\Psi_u(\g^u_1,\dots,\g^u_N)$ we proceed as follows.
\begin{figure}[htb]
        \centering
 \begin{overpic}[scale=0.45]{enlarge2}
\put(10,63) {$x_1^t$}
\put(-8,63) {$\hat x_1^t$}
\put(10,31) {$x_2^t$}
\put(-8,31) {$\hat x_2^t$}
\put(39,21) {$\g_2^t$}\put(55,70) {$\g_1^t$}
\put(83,58) {$y_1^t$}\put(101,58) {$\hat y_1^t$}
\put(83,31) {$y_2^t$}\put(101,31) {$\hat y_2^t$}
\end{overpic}

        \caption{
        Picture of two open contours $\hat\g^t_i:\hat y_i^t\to \hat x_i^t$, $i=1,2$     in the enlarged strip $\cS_v'$.  The paths are obtained by adding straight lines to the corresponding irreducible components $\g_i^t$.  }\label{fig:3}
\end{figure}

We start with $u=t$.
Let $\cS_v'$ denote the infinite vertical strip obtained by enlarging the original strip $\cS_v$ by $(\log L)^2$:    $$\cS_v' = \{x\in\bbZ^2:\, d(x,\cS_v)\leq (\log L)^2\}.$$
Let $\hat x^t_i$ denote the point on the left boundary of $\cS_v'$ which has the same vertical coordinate as $x^t_i$ and let $\hat y^t_i$ denote the point on the right boundary  of $\cS_v'$ which has the same  vertical coordinate as $y^t_i$. Let $\hat\g^t_i$ denote the open contour joining $\hat x^t_i$ and $\hat y^t_i$
obtained by connecting $\hat x^t_i$ and $x^t_i$ by a straight line, then using $\g^t_i$ from $x^t_i$ to $y^t_i$ and then connecting $y^t_i$ and $\hat y^t_i$ by a straight line; see Figure \ref{fig:3}.
This defines  a set of ordered, non-crossing paths $\hat\g^t_i$, $i=1,\dots,N$ in the strip $\cS_v'$,
all staying above the square $\L'$. For a given choice of $\g^t_1,\dots,\g^t_N$, we define the potential:
 \begin{align}\label{ub230}
 \Psi_t(\g^t_1,\dots,\g^t_N) := \Phi_{L',\infty}(\hat\g^t_1,\dots,
\hat\g^t_N)\,,
\end{align}
where $L'=\lfloor(1-\d)L+ (\log L)^2\rfloor$ is half the width of  the strip $\cS_{v}'$, and $\Phi_{L',\infty}$ is defined in \eqref{ub4a1}.

The potentials $\Psi_u(\g^u_1,\dots,\g^u_N)$,
for $u=\ell,b,r$ are defined in the very same way, with the obvious modifications.
\begin{lemma}\label{lemmapsi}
Let $\Psi_{\L_M}$ denote the potential from \eqref{ub21}. There exists $\b_0,C>0$ such that: for any choice of $\g_1,\dots,\g_N$ in \eqref{ub20} with
$\g_1\subset \L_{L^2/2}$, for any $\b\geq \b_0$ one has
\begin{align}\label{ub231}
& |\Psi_{\L_M}(\g_1,\dots,\g_N)-\sum_u\Psi_u(\g^u_1,\dots,\g^u_N) |\nonumber \\
& \quad \qquad \leq C \sum_{i=1}^{N+1}( |\h_i^{t,\ell}|+|\h_i^{\ell,b}|+|\h_i^{b,r}|+| \h_i^{r,t}|) + C(\log L)^3
\end{align}
\end{lemma}
\begin{proof}
We are going to use the properties of the potentials listed in Lemma \ref{lem:dks}.
In particular, we use the fact that for $\b$ large enough, for any $\G\subset \bbZ^2$, any $\l>0$ one has
\begin{align}\label{ub2310}
\textstyle
\sumtwo{V\subset \bbZ^2:}{V\cap \G\neq \emptyset,\,d(V)\geq \l}\sup_{{U_+,U_-}}|\varphi_{U_+,U_-}(V)|\leq
 C|\G|\,e^{-\l}
\end{align}
for some constant $C>0$.
In the potential $\Psi_{\L_M}$ one has a sum over subsets $V\subset \L_M$, while the potential $ \Psi_u$
contains sums over $V$ in the corresponding strips of width $2L'$.  Since we assume $\g_1\subset \L_{L^2/2}$, one has that $d(\g_1,\L_M^c)>L^2/4$ and
therefore adding all $V$'s which are not contained in $\L_M$ does not change the value of $\Psi_{\L_M}(\g_1,\dots,\g_N)$ by more than a constant.
Similarly, using the fact that there are $N=O(\log L)$ contours and that $\g^t_i$ is at distance at least $\l=(\log L)^2$ from the complement of $ \cS'_v$,
when we compute $\Psi_t$, we may remove the constraint that $V\subset \cS'_v$ at the cost of an additive term $O((\log L)^3)$. Indeed,
separating  the contribution from the straight pieces in $\hat \g^t_i$, and observing that $\max_i |\g^t_i|\leq CL^2 $ (since all contours belong to $\L_M$, with $M=L^2$) one has that the sum over all $V\not\subset \cS'_v$ at distance less than $1$ from $\cup_{i=1}^N\hat\g_i^t$
contributes at most
$$
C N L^2 e^{-(\log L)^2} + CN(\log L)^2\leq C(\log L)^3 .
$$
The same applies to all $\Psi_u$, $u\in\{t,\ell,b,r\}$.
The same reasoning shows that the sum over all $V$'s such that $V$ intersects
both $\g^u_i$ and $\g^v_j$, for arbitrary $i,j$ is at most a constant if  $u\neq v$.
It remains to deal with the contribution from all the $V$'s which intersect some corner term $\h_i^{u,v}$. By the rough bound \eqref{ub2310} these can be estimated by $C|\h_i^{u,v}|$. Putting together these facts one arrives at \eqref{ub231}.   \end{proof}

From \eqref{ub21}, if $\g_1\subset \L_{L^2/2}$, then
 Lemma \ref{lemmapsi} implies for $\b$ large enough:
\begin{align}\label{ub25}
\wt \bbP\big( \cap_{i=1}^N\sC_{\g_i,i}\big) &\leq
\textstyle \exp{\big(-\frac12\b\sum_{i=1}^N(|\h_i^{t,\ell}|+|\h_i^{\ell,b}|+|\h_i^{b,r}|+| \h_i^{r,t}|)+ C(\log L)^3\big)}\times \nonumber\\
& \qquad \textstyle
\times\prod_{u}\exp{\big(-\b\sum_{i=1}^N|\g_i^u| + \Psi_u(\g_1^u,\dots,\g_N^u)\big)},
\end{align}
Let us now go back to \eqref{ub20}. Using a very rough bound one can easily obtain
\begin{align}\label{ub24}
\wt \bbP( \g_1\not \subset \L_{L^2/2} )\leq e^{-L^2}.
\end{align}
Indeed, write the expansion \eqref{ub2a1}
 with only one contour and estimate the decoration term $|\psi_\L(\g_1)|\leq c_\b|\g_1|$, with a constant $c_\b>0$ that vanishes as $\b\to\infty$, and then use a simple Peierls' argument together with the fact that $\g_1\not \subset \L_{L^2/2}$ implies $|\g_1|\geq L^2/2$.

From \eqref{ub24} and  \eqref{ub25}, summing over all choices of the points
$$(x,y)= \{(x_i^u,y_i^u),\, i=1,\dots,N; u=t,\ell,b,r\},$$ one has
that up to the additive error term $e^{-L^2}$, $\wt\bbP(A(\d,K))$ is upper bounded by
\begin{align}\label{ub26}
\sum_{(x,y)} \Big(\prod_{i=1}^N\Theta(x_i^t,y_i^\ell)\Theta(x_i^\ell,y_i^b)\Theta(x_i^b,y_i^r)\Theta(x_i^r,y_i^t)\Big)\prod_u \cZ_u(x^u,y^u) ,
\end{align}
where
\begin{align}\label{ub27}
\cZ_u(x^u,y^u) := \sum_{\g_1^u,\dots,\g_N^u}\exp{\textstyle \big(-\b\sum_{i=1}^N|\g_i^u| +\Psi_u(\g_1^u,\dots,\g_N^u)\big)},
\end{align}
and
\begin{align}\label{ub28}
\Theta(x_i^u,y_i^v) := e^{C(\log L)^3}\!\!\!\!\!\sum_{\eta:\,x_i^u\to y_i^v}\exp{(-\textstyle \frac12\b|\eta|)}.
\end{align}
The sum in \eqref{ub27} ranges over all open contours $\g_i^u:y_i^u\to x_i^u$ such that  $\g_{i}^u,\g_{j}^u $ do not cross for $i\neq j$ and such that $\g_{i}^u$ is more internal (closer to $\L'$) than $\g_{j}^u$ for $i>j$. Since we are doing an upper bound,  we may neglect the constraint that $\g_i^u$ does not cross the boundary of $\L'$. The sum  in \eqref{ub28}
ranges over all paths from $x_i^u\to y_i^v$.
The following lemma summarizes the main estimate we need.
\begin{lemma}\label{prozs}
For any $u$, uniformly in the choice of the points $x^u,y^u$, one has
\begin{align}\label{ub29}
\cZ_u(x^u,y^u) \leq \exp {\big(-2\b \t_\b(0)N L (1-\d) (1+o(1))\big)}  .
\end{align}
\end{lemma}
Let us conclude the proof by assuming the validity of Lemma \ref{prozs}.
From \eqref{ub28} one has
$$
\sum_{x_i^u,y_i^v}
\Theta(x_i^u,y_i^v) \leq e^{C(\log L)^3},
$$
for some new constant $C$. Therefore, one has the upper bound
$$
 \sum_{(x,y)} \Big(\prod_{i=1}^N\Theta(x_i^t,y_i^\ell)\Theta(x_i^\ell,y_i^b)\Theta(x_i^b,y_i^r)\Theta(x_i^r,y_i^t)\Big)
 \leq e^{C(\log L)^4}.
 $$
From \eqref{ub20}-\eqref{ub26}, using the uniform bound \eqref{ub29} for each $u$, one has
\begin{align}\label{ub30}
\wt \bbP( A(\d,K)) \leq e^{C(\log L)^4}\exp {\big(-8\b \t_\b(0)N (1-\d)L  (1+o(1))\big)}.
\end{align}
Since $N=\frac1{4\b}\log L(1+o(1))$ the conclusion \eqref{ub2} follows.

\subsection{Proof of Lemma \ref{prozs}}\label{mainproof}
The core of the proof is the monotonicity argument of Theorem \ref{th:m1} that allows us to consider each of the $N$ contours separately; see Section \ref{monostate}.
To be able  to apply this we first need to reformulate the problem in terms of $SOS$ contours.

Without loss of generality we assume that $u=t$.

Let $\hat x_i^t,\dots,\hat y_i^t$ denote the points on the boundary of $\cS_v'$ as defined before
\eqref{ub230}, and call $a_{N-i+1}$ the vertical coordinate of $\hat x_i^t$ and $b_{N-i+1}$ the vertical coordinate of $\hat y_i^t$, $i=1,\dots,N$. Let $\cZ(a_1,\dots,a_N;b_1,\dots,b_N; L')$, $L'=(1-\d)L+(\log L)^2$, denote the partition function of the $N$ contours in the strip $\cS_v'$ as defined in Lemma \ref{lem:infvol}.

We claim that
\begin{align}\label{ub31}
\cZ_t(x^t,y^t)\leq e^{C(\log L)^3} \,
\cZ(a_1,\dots,a_N;b_1,\dots,b_N; L').
\end{align}
Let us first conclude the proof of Lemma \ref{prozs} assuming the validity of the estimate \eqref{ub31}. From \eqref{ub31} and
Theorem \ref{th:m1} we can bound $\cZ_t(x^t,y^t)$ from above by a product of partition functions of a single contour:
\begin{align}\label{ub32}
\cZ_t(x^t,y^t)\leq e^{C(\log L)^3} \,
\prod_{i=1}^N\cZ(a_i;b_i; L').
\end{align}
The surface tension bound \eqref{eq:surftens2} then implies the desired estimate
\eqref{ub29}.

To conclude the  proof of Lemma \ref{prozs}, it remains to prove \eqref{ub31}.
To this end, observe that by the expansion \eqref{ub4a1}, one has
\begin{align}\label{ub34}
\cZ(a_1,\dots,a_N;b_1,\dots,b_N; L')
=\sum_{\hat \g_1,\dots,\hat\g_n}\exp{\textstyle \big(-\b\sum_{i=1}^N|\hat\g_i| +\Phi_{L',\infty}(\hat\g_1,\dots,\hat\g_N)\big)},
\end{align}
where the sum ranges over all collections of non-crossing contours $\hat\g_i:\hat x_i^t\to \hat y_i^t$.
Let us restrict this summation to paths of the form $\hat\g_i = \hat\g^t_i$, i.e.\ paths which have a straight line from $\hat x_i^t$ to $x_i^t$, a regular path $\g_i^t:x_i^t\to y_i^t$, and a straight line from $y_i^t\to \hat y_i^t$; see Figure \ref{fig:3}. By summing over the regular parts $\g_i^t$ and using $|\hat\g^t_i|=|\g^t_i|+2(\log L)^2$ one has
\begin{align}\label{ub35}
&\cZ(a_1,\dots,a_N;b_1,\dots,b_N; L')\nonumber \\
&\qquad \geq \sum_{\g^t_1,\dots,\g^t_n}
\exp{\textstyle \big(-\b\sum_{i=1}^N|\g^t_i| +\Phi_{L',\infty}(\hat\g^t_1,\dots,\hat\g^t_N) - 2\b N(\log L)^2}\big),
\end{align}
By the definition \eqref{ub230}, one has $\Phi_{L',\infty}(\hat\g^t_1,\dots,\hat\g^t_N) = \Psi_t(\g^t_1,\dots,\g^t_N)$. Therefore, using $N\leq (4\b)^{-1}\log L$, we conclude
\begin{align}\label{ub36}
\cZ(a_1,\dots,a_N;b_1,\dots,b_N; L')
\geq \cZ_t(x^t,y^t)\,e^{-C(\log L)^3}.
\end{align}
This ends the proof of \eqref{ub31}.

\bibliographystyle{plain}
\bibliography{sos.bib}

\end{document}

config: arxiv


TITLE: Prove that in acute-angled triangle middle point of a sector lies at a certain line
QUESTION [0 upvotes]: Segments AD, BE are heights of acute-angled triangle ABC. Point M is middle point of AB segment. Points P, Q are symetrical to point M under respectively line AD, BE. Show that center point of DE segment lies on straight PQ.
Here's a picture for better refrance https://puu.sh/recsD.png (I've used geogebra to create a small applet). Points G and H are cross points of AD with PM and BE with MQ. Point F is middle point of DE (probably PQ as well)
I've been trying to prove it for past few hours without much progress.
I figured that:

PM || BC, QM || AC
angle PMQ = angle ACB
triangles CEB and CDA are simillar.

Any help appreciated :)

REPLY [0 votes]: There are several variations, but let me mention one. Segment $MH$ is orthogonal to $BE$ and since $AE$ is orthogonal $BE$ the two segments $MH$ and $AE$ are parallel. Since $M$ is a midpoint of $AB$, segment $MH$ is half of $AE$ in length. Analogously, $MG$ is parallel to $BD$ and since $M$ is a midpoint of $AB$, point $G$ is a midpoint of $AD$.
Look at quadrilateral $MGFH$. Segment $FG$ is a midpoint-segment in triangle $AED$ and is therefore parallel to $AE$ and half of its length. Thus $MH$ is parallel to $GF$ and $MH = GF$. Hence $MGFH$ is a parallelogram. Then, $HQ=MH = GF$ and since $MH$ is parallel to $GF$, segment $HQ$ is also parallel to $GF$. Thus $HGFQ$ is a parallelogram and therefore $FQ$ is parallel to $HG$.
Segment $HG$ is a midpoint segment in triangle $MPQ$ and is therefore parallel to $QP$. Now, since $QF$ and $QP$ are both parallel to $HG$ and since there is only one line passing through $Q$ and is parallel to $HG$, lines $QF$ and $QP$ coincide and thus points $Q, F, P$ are collinear.


TITLE: Point in Polygon algorithm from the viewpoint of a robot
QUESTION [6 upvotes]: I've come across the following puzzle:

You're on an island, on which there is
  a fence (which is a simple closed
  contour). You need to determine
  whether you're inside or outside the
  fence.

Now if you had the function defining the contour as well as the point you're in (e.g. you have a GPS), you could calculate the winding number. If you could climb over the fence, you could use ray casting. Both methods are described here.
However, I've come across a claimed solution that presumably does not require any of them:
Traverse the fence - say by keeping the fence stuck to your left side, until you've come back where you started, (assuming you can mark your start position). Repeat the process where the fence is always 1 meter to your left, orthogonally (assuming you can maintain orthogonality, maintain the 1 meter distance and the fence is always wide enough for you to maintain it).
The claim is - if the second trip (the one walking 1 meter away from the fence) took more time (assuming you can measure time and maintain the same exact speed throughout both trips), you're in the exterior. otherwise you're in the interior.
I haven't been able to prove this, and I'm not even sure it's right (couldn't find a counterexample, though).
Any thoughts ?

REPLY [2 votes]: Lookj up "level sets".
The wikipedia page "Point in Polygon" talks about algorithms that can be used when the polygon's coordinates are known.  The proposed solution of the length of a path following the fence along the fence (call it $d_0$) and a path following along the fence but maintaining a constant distance of $x=1$ meter (call it $d_1$) will work for a robot that can do the tasks you're asking of it.  However, the answer of the difference in length being $2 \pi \sim 6.28$ meters would only apply if the fence is perfectly circular.
Given a map or diagram of the fence, generate multiple contours or level sets of points which are a constant distance from the fence.  You'll end up with something that looks like a contour map or topographical map that the U.S. Geological surveys generates. Notice that for each distance $x$ (up to a certain limiting value), the level sets for $d_x$ may contain points inside the fence as well as outside the fence.  Once $x$ is greater than the radius of the circle, the level sets for $d_x$ such that $x \gt r$ will only contain points outside the circle.  For fences with concavities (like a pinched figure 8), the inner level set may break up into multiple non-connected paths.
If the fence is square, width edge length $2r$, then the close-fence contour $d_0$ will be $4 \times 2r = 8r$, whereas the 1-meter level set will be

$4 \times (2r) + 4 \times (\frac{1}{2} \pi) = 8r + 2 \pi $ if the robot's path is 1-meter outside the fence (which consists of the edges translated outward a distance of 1-meter, and of quarter-circle arcs at each of the corner, as correctly pointed out by Mark Bennet's comment below).
$4 \times (2r-2) = 8r - 4$ if the robot path is 1-meter inside the fence.

Thus for square, non-convex, and pretty much any noncircular fence, the level-set path one meter of the fence will not be $2 \pi \sim 6.28$ meters different from the level-set path of distance $0$ from the fence.
The generalization, however, will still apply that the level set path of distance $x$ away from the fence will be smaller ($d_x \lt d_0$) if the robot follows the level set path within the fence, vs. larger if the robot follows the level set path outside the fence ($d_x \gt d_0$).


TITLE: How is there moment for a group of forces which its resultant equals zero?
QUESTION [0 upvotes]: I am a beginner in mechanics, and when I was introduced to moments there was a theorem that says "if a set of forces acting on a rigid body has a resultant, then the algebraic sum of these forces about a certain point is equal to the moment of the resultant about this point" so what I understood is if the resultant of forces is zero then the moment of these forces should be zero too by applying the previous theorem and also because M=rxR so if R equals zero then the moment equals zero too, but then I learned couples, in couples the resultant equals zero but the moment doesn't equal zero how is this if M=rxR then moment should be zero. I hope that you understand my question and I will be thankful for anybody who answers it.

REPLY [0 votes]: Lets try to get clear.  The vector sum of the forces acting on an object determines the rate of change of the linear momentum. This change can be expressed as MA, where (M) is the total mass and (A) is the vector acceleration of the center of mass. Similarly, the vector sum of the torques about a point within an object determines the rate of change of angular momentum about that point. (This gets more complicated.) I'm assuming that when you refer to a moment, you are talking about a torque.  A torque can be best defined as the work per unit angle of rotation (as in Joules/radian) that can be done by a force acting in a manner that tends to cause a rotation. In the simplest case of an object rotating about a fixed axle, this reduces to the radius measured (perpendicular) from the axle to the point of application of the force multiplied by the tangential component of the force. In that case, the sum of torques about that axle determines the angular acceleration about that axle.  Each torque and the angular acceleration are represented by vectors along the axle. Sometimes, you may want to use the vector sum of torques about the center of mass to find the angular acceleration about the center of mass. In a static system (with no acceleration) the sum of torques about any point must be zero.