text:
\section{Acknowledgements}
Riley Borgard, Haley Duba, Chloe Makdad, Jay Mayfield, and Randal Tuggle were supported by the National Science Foundation through the REU award \#1457443.
Steven Harding and Eric Weber were supported by the National Science Foundation and the National Geospatial-Intelligence Agency under award \#1830254.  Eric Weber was also supported under award \#1934884.


TITLE: Weak convergence in $L^2[0,1]$
QUESTION [3 upvotes]: In $\ell^2$, a sequence converges weakly if it is bounded and each component converges. My question is: How can I think about weak convergence in $L^2[0,1]$?

REPLY [1 votes]: The space $L^2([0,1])$ equipped with the inner product
$$\langle f,g\rangle:=\int^1_0f(x)\overline{g(x)}\,dx$$
is a Hilbert space. By definition of weak convergence we say a sequence $(f_n)\subset L^2([0,1])$ converges weakly to $f\in L^2([0,1])$ whenever
$$\lim_n\langle f_n,g\rangle=\langle f,g\rangle\hspace{0.2cm}\text{for all}\hspace{0.2cm}g\in L^2([0,1])$$
On the other hand the spaces $l^2(\mathbb{N})$ and $L^2([0,1])$ are isometric isomorphic to each other. To see this note that if $(e_n)\subset L^2([0,1])$ is a complete orthonormal basis for $L^2([0,1])$ then every $f\in L^2([0,1])$ admits a representation of the form
$$f=\sum_{n=1}^{\infty}\langle f,e_n\rangle e_n\Rightarrow ||f||^2_{L^2}=\int_0^1f(x)\overline{f(x)}\,dx=\langle f,f\rangle=\sum^{\infty}_{n=1}|\langle f,e_n\rangle|^2:=||f_n||^2_{l^2}$$
implies $(f_n)\in l^2(\mathbb{N})$ where $f_n:=\langle f,e_n\rangle$. Clearly the operator $A:L^2([0,1])\mapsto l^2(\mathbb{N})$ defined by $Af:=(f_n)$ is a linear bounded operator which preserves the norms (as we can see for the above relation). Now to come back to the initial statement that if $(f^k_n)$ is a bounded sequence for each $k$ and for each $n$ we have $|f^k_n-f_n|\to 0$ as $k\to\infty$ then $(f^k_n)$ converges weakly to $(f_n)\in l^2(\mathbb{N})$. Again by definition $(f^k_n)$ converges weakly to $(f_n)$ whenever $$\lim_k\langle f^k,g\rangle=\langle f,g\rangle$$
where the inner product for $l^2(\mathbb{N})$ is defined by
$$\langle f,g\rangle:=\sum_nf_n\overline{g_n}$$
Boundedness of each $(f^k_n)$ implies $(f^k_n)\in l^2(\mathbb{N})$ for each $k$ so their inner product with any other element $g$ of $l^2(\mathbb{N})$ is finite and well defined. On the other hand component wise convergence, $|f^k_n-f_n|\to 0$ as $k\to\infty$ for each $n$, implies
$$|\langle f^k,g\rangle-\langle f,g\rangle|=|\langle f^k-f,g\rangle=\Big|\sum_n(f^k_n-f_n)\overline{g_n}\Big|\leqslant \sum_n|f^k_n-f_n||\overline{g_n}|\to 0$$
as $k\to\infty$. Therefore the same definition for both spaces is valid.


TITLE: Fibonacci coding
QUESTION [4 upvotes]: Show that every positive integer can be written as a sum of distinct terms of the Fibonacci sequence. (The Fibonacci sequence $\{F_n\}_n$ is defined by $F_0=0, \ F_1=1,$ and $F_{n+1}=F_n+F_{n-1}, \ n\geqslant 1$.)
My proof: We'll prove that statement via the strong mathematical induction.
We see that for $n=1$ it is true since in that case $n=F_1$ or we can take $n=F_0$.
Suppose that our statement is true for $1\leqslant n \leqslant k$ and let's prove this for $n=k+1$.
Then $\exists m\in \mathbb{N}: \ F_m\leqslant k+1<F_{m+1}$.
If $k+1=F_m$ then we are done.
If $F_m<k+1<F_{m+1}$ then $0<(k+1)-F_m\leqslant k$ and here we can apply the assumption of our induction and also $0<(k+1)-F_m<F_{m+1}-F_m=F_{m-1}$ and this guarantees that there is no more $F_{m}$ in $(k+1)-F_m$.
Thus, we have proven out statement.
Is my reasoning correct?

REPLY [3 votes]: Your argument is correct in essence.  I'd say something like: "If $F_m < k+1 < F_{m+1}$, then $k+1 = F_m+j$, where $0 < j < F_{m+1}-F_m = F_{m-1}$.  By strong induction, $j$ can be expressed as the sum of distinct Fibonacci terms, none of which can be as great as $F_{m-1}$, which is less than $F_m$, $k+1$ can likewise be expressed as a sum of distinct Fibonacci terms."  But that's a matter of personal preference.


TITLE: Asymptotic estimation of $A_n$
QUESTION [10 upvotes]: Let $A_n$ represent the number of integers that can be written as the product of two element of $[[1,n]]$.
I am looking for an asymptotic estimation of $A_n$.
First, I think it’s a good start to look at the exponent $\alpha$ such that :
$$A_n = o(n^\alpha)$$
I think we have : $2 \leq \alpha $. To prove this lower bound we use the fact that the number of primer numbers $\leq n$ is about $ \frac{n}{\log n}$.
Hence we have the trivial lower bound (assuming $n$ is big enough) :
$$ \binom{ E(\frac{n}{\log n})}{2} = o(n^2)$$
Now is it possible to get a good asymptotic for $A_n$ and not just this lower bound ?
Is what I’ve done so far correct ?
Thank you !

REPLY [2 votes]: The exact asymptotic isn’t known, but Kevin Ford astutely worked out the exact order of magnitude, not too long ago.  So the only unknown is the constant in front (or perhaps even whether such a constant exists).
From the answer at https://math.stackexchange.com/a/1336590/30402, the value of $A_N$ satisfies upper and lower bounds of the form
$$
A_N \asymp \frac{N^2}{(\log N)^c(\log\log N)^{3/2}}$$ where $$c=1-\frac{(1+\log \log 2)}{\log 2}.
$$


TITLE: Prove that (a+bc) + (a'(b'+c')) = 1
QUESTION [1 upvotes]: Im trying to prove the boolean expression (a+bc) is the complement of (a'(b'+c')). To do that I need to prove that (a+bc) + (a'(b'+c')) = 1, however I only get as far as:
(a + bc + a')(a + bc + b' + c')
(a + a' + bc)(a + bc + b' + c')
(1 + bc)(a + bc + b' + c')
1(a + bc + b' + c')

What property can I use to reduce the RHS?

REPLY [0 votes]: $$a \lor bc \lor \overline a(\overline b \lor \overline c)=1$$
I use this rule: $a=a \lor a \overline b=a \lor a \overline c$.
$$[a \lor bc \lor \overline a \overline b \lor \overline a \cdot \overline c] \lor a \overline b \lor a \overline c =1 $$
You should remember that $ x \lor \overline x = 1$.
$$ a \lor \overline b \lor \overline c \lor bc =1$$
$$ a \lor \overline b \lor \overline c \lor bc \lor \overline b c =1$$
$$ a \lor \overline b \lor \overline c \lor c =1$$
$$ a \lor \overline b \lor 1 =1$$
So, this function is always true.


TITLE: Are pullbacks on singular cohomology unique on the nose?
QUESTION [6 upvotes]: Let $C$ be the category of even-dimensional connected closed oriented topological manifolds/orientation-preserving continuous maps and $D$ be the category of finite-dimensional graded $\mathbb{Q}$-algebras.
We have a functor $H:C\rightarrow D$ given by singular cohomology.
Let us call a choice of an isomorphism of graded $\mathbb{Q}$-algebras $H(M)\otimes H(M')\rightarrow H(M\times M')$ for every $M$, $M'\in Obj(C)$ a Kunneth system. Projections $M\times M'\rightarrow M$, $M'$ define a Kunneth system which is functorial with respect to $H$ (Kunneth theorem).
For a contravariant functor $G:C\rightarrow D$ such that the induced map on objects is equal to one induced by $H$, it is meaningful to ask if the Kunneth system above is functorial with respect to $G$. The question: does there exist a contravariant functor $G:C\rightarrow D$ such that

$G$ coincides with $H$ on the level of objects but not on the level of morphisms;
for any non-negative even integer $i$, the contravariant functors from $C$ to $\mathbb{Q}$-vector spaces obtained by composing either $G$ or $H$ with the "$i$-th graded piece" functor are equal;
the Kunneth system above is functorial with respect to $G$?

P.S.: this question is inspired by this question.

REPLY [1 votes]: The answer to the question as posed is no, if I have understood the setup correctly.
It suffices to show that a contravariant functor $G:C\to D$ satisfying your second and third bullet points must agree with $H:C\to D$ on morphisms. This would follow if we can show that for every $a\in H^i(M)=G^i(M)$ with $i$ odd and for every morphism $f:N\to M$ in $C$ we have $H(f)(a)=G(f)(a) \in H^i(N)=G^i(N)$.
To see this, take a second class $b\in H^j(M')=G^j(M')$ with $j$ odd, and a second morphism $g:N'\to M'$. Observe that by naturality of the Kunneth isomorphisms for $G$ and $H$ and the fact that $a\times b\in H^{i+j}(M\times M')$ with $i+j$ even, so that $H(f\times g)(a\times b)=G(f\times g)(a\times b)$, we have that
$$G(f)(a)\otimes G(g)(b)=H(f)(a)\otimes H(g)(b) \in H^i(N)\otimes H^j(N') = G^i(N)\otimes G^j(N').$$
The tensor product is over the rationals. We may take $N'=M'$ and $g=\operatorname{Id}_{M'}$, the identity map. By functorialty $G(g)(b)=b=H(g)(b)$, and so
$$
G(f)(a)\otimes b=H(f)(a)\otimes b
$$
which by bilinearity of tensor products gives that
$$(G(f)(a)-H(f)(a))\otimes b = 0.$$
Choosing $b$ to be nonzero, this implies that $G(f)(a)=H(f)(a)$, as claimed.


text:
\begin{document}

\begin{titlepage}

\vspace*{0.2cm}

\renewcommand{\thefootnote}{\star}
\begin{center}

{\LARGE\bf  An 8-dimensional Taub-NUT-like hyper-K\"ahler metric in harmonic superspace formalism}

\vspace{2cm}

{\Large A.V. Smilga} \\

\vspace{0.5cm}

{\it SUBATECH, Universit\'e de
Nantes,  4 rue Alfred Kastler, BP 20722, Nantes  44307, France. }

\end{center}
\vspace{0.2cm} \vskip 0.6truecm \nopagebreak

   \begin{abstract}
\noindent
Using the harmonic superspace  formalism, we find the metric of a certain 8-dimensional manifold. This manifold is not compact and represents an  8-dimensional generalization of the Taub-NUT manifold. Our conjecture is that the metric that we derived is equivalent to the known metric possessing a discrete $Z_2$ isometry, which may be obtained from the metric describing the dynamics  of four BPS monopoles by Hamiltonian reduction.
   \end{abstract}

\end{titlepage}

\setcounter{footnote}{0}

\setcounter{equation}0
\section{Introduction}

Hyper-K\"ahler (HK) manifolds admit a triple of covariantly constant quaternionic complex structures,
\be
\lb{3-structures}
\nabla_M I_{NL}^{p = 1,2,3} \ =\ 0,  \qquad (I^p)_M{}^N  (I^q)_N{}^L \ =\ -\delta^{pq} \delta_M^L + \varepsilon^{pqr} (I^r)_M{}^L  \,.
 \ee
 Their dimension is an integer multiple of 4,
 \be
 \lb{D=4n}
  D \ =\ 4n \,.
  \ee
  It is well known that the supersymmetric sigma models with HK target spaces enjoy extended supersymmetries. This is true for field-theory models \cite{AG-F} and also for supersymmetric quantum mechanical (SQM) models where the variables and supervariables depend only on time, but not on spatial coordinates.

  Such HK models including $4n$ real  ({\bf 1}, {\bf 2}, {\bf 1}) supermultiplets \footnote{We use the notation introduced in \cite{Toppan}:  the first number counts the dynamical real bosonic variables, the second number counts the dynamical real fermionic variables and the third number counts the bosonic auxiliary variables.} possess ${\cal N} = 8$ supersymmetry, and the models including $2n$ chiral  ({\bf 2}, {\bf 2}, {\bf 0}) supermultiplets possess ${\cal N} = 4$ supersymmetry. In this article, we will only discuss the second type of models.

  The ${\cal N} = 4$ SQM superfields depend on time, on $\theta_{j=1,2}$ and on $\bar \theta^j$. A generic such superfield,
  \be
  \lb{Phi-gen}
  \Phi(t, \theta_j, \bar \theta^j) \  = \ \phi(t) + \theta_j \psi^j(t) + \ldots,
   \ee
   involves 16 complex component variables. And  ${\cal N} = 8$ superfields depending
   on $t, \theta_{j\alpha}, \bar \theta^{j\alpha}$ involve in the expansion 256 complex components.
   But this description is not convenient. In particular, it is difficult in the ordinary language to find the constraints to be imposed on a generic superfield to make the constrained superfield an irreducible representation of the supersymmetry algebra.

   This can be done, however, in the {\it harmonic superspace} (HSS) formalism
   \cite{HSS}. In \cite{L+4-HK}, this formalism was applied to describe  the
   ${\cal N}=8$ supersymmetric sigma models living on HK target spaces.  The ${\cal N} = 4$ models living on HK manifolds were described in
   \cite{DI}.\footnote{The formalism developped in Ref. \cite{DI} also describes  the so-called HKT manifolds, where the complex structures are covariantly constant with respect not to the ordinary Levi-Civita connections, but to certain special {\it Bismuth connections} including nonzero torsions. But we will be interested only in the HK manifolds in this paper.}
   Choosing a harmonic prepotential ${\cal L}^{+4}$, it is possible to find the corresponding HK metric.

   However, in contrast to K\"ahler manifolds, where the metric is derived from the K\"ahler potential by simple differentialtion, $h_{j\bar k} = \partial_j \partial_{\bar k} {\cal K}(z^n, \bar z^n)$,  the hyper-K\"ahler problem is much more difficult. To derive the metric from ${\cal L}^{+4}$, one has to solve a set of rather complicated harmonic differential equations. One can always do so
 perturbatively in the neighbourhood of a given point on the manifold, but only in rare cases one can find an analytic solution to these equations. In \cite{L+4-HK}, this was explicitly done only for two 4-dimensional HK manifolds: the Taub-NUT manifold and the Eguchi-Hanson manifold.

 In this paper, we explicitly derive in this way a particular 8-dimensional HK metric. This metric is not compact and represents a generalization of the Taub-NUT metric.

\section{One-dimensional harmonic superspace.}
\setcounter{equation}0

With the main purpose to establish notations, we present here the very essentials of the harmonic superspace formalism. For more  details, which are necessary for understanding, the reader is addressed to the monography \cite{HSS}, to the paper \cite{IL}, where the HSS technique, invented originally to deal with extended supersymmetric field theories, was adapted for SQM systems, and to Chap. 7.4 and Chap. 12 of recent \cite{glasses}.

The harmonic superspace technique  capitalizes on the presence of the $R$-symmetry
that rotates complex supercharges and the associated fermion variables. For ${\cal N} = 4$ systems, this symmetry is $U(2)$ and its relevant for the following part is $SU(2)$.  The basic idea is to {\it extend} the ${\cal N} = 4$ superspace $(t; \theta_j, \bar \theta^j)$ by adding  extra commuting coordinates $u^{+j}$ parameterizing this $SU(2)$ group. $u^{+j}$ is a complex spinor of unit length:
 \be
\lb{norm-u}
u^{+j} \overline{u^{+j}} \ \stackrel {\rm def}= \ u^{+j}u^-_j \ = \ 1 \,.
 \ee
 The index $j$ is lowered and lifted by $\varepsilon_{jk} = -\varepsilon^{jk}$ with the convention $\varepsilon_{12} = 1$. The matrix
    \be
    \lb{u+u-matr}
    U \ =\ \left( \begin{array}{cc} u^+_1   &  u^-_1 \\ u^+_2 & u^-_2 \end{array} \right)
    \ee
    belongs to $SU(2)$ so that
 the property
  \be
  \lb{uu-uu}
  u^+_j u^-_k - u^-_j u^+_k \ =\ \varepsilon_{jk}
   \ee
   holds.

 Introduce now the odd coordinates $\theta^\pm = \theta^j u^\pm_j, \ \ \ \bar \theta^\pm = \bar \theta^j u^\pm_j$. They are  $SU(2)$ invariants.
  The harmonic superspace is parameterized by the coordinates
   \be
\lb{full-harm}
 t; \ \theta^+, \bar \theta^+, \theta^-, \bar \theta^-; \ u^\pm_j \, .
 \ee
 One can define a subspace of this superspace (called {\it Grasmann-analytic} superspace) parameterized by $(t_A;\ \theta^+,
\bar \theta^+; \ u^\pm_j )$,
where
\be
\lb{anal-time}
t_A \ =\ t + i(\theta^+ \bar \theta^- + \theta^- \bar \theta^+)
 \ee
is the so-called {\it analytic time}. The point is that the Grassmann-analytic superspace is invariant under supersymmetry transformations.\footnote{The analogy with familiar chiral superspaces is straightforward.}.
It is also invariant under the  special {\it pseudoconjugation}  operation---a combination of the ordinary complex conjugation and the following transformation of the harmonics: $u^+_j  \to  u^-_j, \ u^-_j  \to  - u^+_j$. The pseudoconjugation acts as
\begin{equation}\label{harm-tilde}
\widetilde{\theta^\pm}=\bar\theta^\pm\,,\qquad \widetilde{\bar\theta^\pm}=-\theta^\pm\,,\qquad
\widetilde{u^\pm_j}=u^\pm{}^j\,.
\end{equation}

 Introduce the supersymmetric harmonic  derivatives:
 \be
\lb{D-harm}
D^0 &=&  \partial^0 \  +  \ \theta^+ \frac {\partial}{\partial \theta^+ }
+  \bar\theta^+ \frac {\partial}{\partial \bar\theta^+} -  \theta^- \frac {\partial}{\partial \theta^- } -
\bar \theta^- \frac {\partial}{\partial \bar \theta^- }\,, \nn
D^{++} &=& \partial^{++}  +\
\theta^+ \frac {\partial}{\partial \theta^- } +  \bar \theta^+ \frac {\partial}{\partial \bar\theta^- } \,,
\nn
D^{--} &=&  \partial^{--} \ + \
\theta^- \frac {\partial}{\partial \theta^+ } +  \bar \theta^- \frac {\partial}{\partial \bar\theta^+ } \,,
\ee
where
\be
\lb{d-harm}
\partial^0 &=&  u_j^+ \frac {\partial}{\partial u_j^+ } -  u_j^- \frac {\partial}{\partial u_j^- }\,, \nn
\partial^{++} &=& u^+_j \frac {\partial}{\partial u_j^-}\,, \nn
 \partial^{--} &=& u^-_j \frac {\partial}{\partial u_j^+} \,.
  \ee
Both \p{D-harm} and \p{d-harm} satisfy the $su(2)$ algebra.
The operator $D^0$ is called the {\it harmonic charge}. All the harmonic superfields that we are going to deal with have a definite integer value of this charge. The operator $D^{++}$ acting on a superfield of harmonic charge $q$ produces a superfield of  charge $q+2$.  The operator $D^{--}$ acting on a  superfield of charge $q$ produces a  superfield of charge $q - 2$.

Consider a Grassmann-analytic superfield of harmonic charge $+1$:
\be
\lb{q+}
q^+(t_A; \theta^+, \bar \theta^+; u)  = f^+(t_A, u) + \theta^+ \chi(t_A, u) +
\bar \theta^+ \kappa(t_A, u) + \theta^+ \bar \theta^+ A^-(t_A, u), \qquad
 \ee
where $ f^+(t_A, u)$ and $A^-(t_A, u)$ are Grassmann-even variables of \index{harmonic charge}  harmonic charge $+1$ and $-1$, respectively, and $\chi(t_A, u)$ and $ \kappa(t_A, u)$
are Grassmann-odd variables of zero \index{harmonic charge}  harmonic charge.

The absence of the dependence of $q^+$ on $\theta^-$ and $\bar \theta^-$ makes the component expansion short. However, the components now depend not only on time, but represent infinite series in $u^+$ and $u^-$.
A remarkable fact, however, is that one can eliminate almost all the terms in these series by imposing a constraint
  \be
\lb{lin-cons}
  D^{++} q^+ \ =\ 0 \, .
  \ee
A general solution to the constraint \p{lin-cons} is
  \be
\lb{q+sol}
q^+(\zeta, u) \ =\ f^j(t_A)u^+_j  +
\theta^+ \chi(t_A) +
\bar \theta^+ \kappa(t_A)  - 2i \theta^+ \bar \theta^+\dot{f}^j(t_A) u^-_j\, ,
 \ee
  We are left in this case with only two complex bosonic variables $f^j(t_A)$
and two complex fermionic variables $\chi(t_A)$ and $\kappa(t_A)$.

 To describe a sigma model living on a $4n$-dimensional HK manifold, we take $2n$ superfields \p{q+} and impose, instead of the constraints \p{lin-cons}, a nontrivial nonlinear constraint \cite{DI}
  \be
\lb{nonlin-cons}
D^{++} q^{+a} \ =\ {\cal L}^{+3a}  \equiv \Omega^{ab} \, \frac {\partial {\cal L}^{+4}(q^{+c}, u) }{\partial q^{+b}}\, ,
  \ee
  where ${\cal L}^{+4}$ is an arbitrary function of its arguments carrying harmonic charge $+4$ and
  $\Omega^{ab}$ is the symplectic matrix, which can be chosen as
\be
\lb{Omega}
  \Omega^{ab} \ =\ -{\rm diag} (\varepsilon, \ldots, \varepsilon)\,.
  \ee
  The symplectic index $a$ will be lifted and lowered by multiplication over $\Omega^{ab}$ and $\Omega_{ab} = - \Omega^{ab}$.

 In addition, we impose on $q^{+a}$ the pseudoreality constraint
  \be
  \lb{harm-pseudoreal}
  \widetilde{q^{+a}} = - q^+_a \,.
  \ee

 Under these conditions, the invariant action
 \be
 \lb{action}
 S  \ =\  - \frac 18  \Omega_{ab} \int dt du \, d\theta^+ d \bar \theta^+  d\theta^- d \bar \theta^-\, q^{+a} q^{-b},
  \ee
  where $q^{-a} = D^{--} q^{+a}$,
  describes the HK geometry \footnote{For a generic ${\cal L}^{+3a}$ in \p{nonlin-cons}, we would arrive to HKT geometry, but this is outside the scope of this paper.} (the coefficient $-1/8$ is a convenient convention). It enjoys  ${\cal N} = 4$ supersymmetry by construction. The expression \p{action} involves, besides the familiar time and $\theta$ integration, also the untegral over the $SU(2)$ group that the harmonics $u^{+j}$ parameterize. This integral can be done using the properties \p{norm-u} and \p{uu-uu}. If we normalize the group volume $\int \! du$ to 1, we derive
  \be
   \lb{harm-int}
  && \int u^+_j u^-_k  \, du \ = \ \frac 12 \varepsilon_{jk}, \nn
  &&\int u^+_j u^+_k u^-_l u^-_p \, du \ =\
    \frac 16( \varepsilon_{jl} \varepsilon_{kp} +
    \varepsilon_{jp} \varepsilon_{kl}) \, , \qquad {\rm etc.}
    \ee
   The integrals of the monoms including  inequal numbers of $u^+$ and $u^-$ vanish.

  To make contact with geometry, we have to express the action \p{action} in components. To begin with, we substitute the expansion \p{q+} into the constraints \p{nonlin-cons}. We derive for the components
   \be
\lb{cons-comp}
\partial^{++} f^{+ a} &=& \Omega^{ab} \, \frac{\partial {\cal L}^{+4}(f^{+c}, u)}{\partial f^{+b}} \,,  \\
{\cal D}^{++} \chi^a &=& {\cal D}^{++} \kappa^a = 0\,, \lb{ferm} \\
{\cal D}^{++} A^{- a} &=&  -2i \dot{f}^{+ a} + \ \Omega^{ab} \frac
{\partial^3 {\cal L}^{+4}(f^{+e}, u)}{\partial f^{+b} \partial f^{+c}  \partial f^{+d}}\,\kappa^c\chi^d\,,
\lb{A-}
\ee
where the action of the covariant harmonic derivative ${\cal D}^{++}$
on any symplectic vector $X^a$ is defined as
       \be
       \lb{D++def-HKT}
{\cal D}^{++} X^a \ =\     \partial^{++} X^a   -  \,\Omega^{ab} \frac {\partial^2 {\cal L}^{+4}}{\partial f^{+b}\partial f^{+c}} \, X^c \,, \nn
{\cal D}^{++} X_a \ =\     \partial^{++} X_a   +  \,\Omega^{cb} \frac {\partial^2 {\cal L}^{+4}}{\partial f^{+a}\partial f^{+b}} \, X_c \,.
     \ee
     The constraint \p{harm-pseudoreal} implies  that
  \be
   f^{+a}(t,u) \ =\ x^{ja} u^+_j + \ {\rm higher\ harmonic\ terms}
    \ee
  with $x^{ja}$ satisfying the pseudoreality condition \p{pseudoreal-V}. As explained in the Appendix, this means that $x^{ja}$ can be traded for $4n$ real coordinates $x^M$.

  To find the metric, we should resolve the equation \p{cons-comp} for $f^{+a}$, then the equation  \p{A-} (with the fermion terms suppressed) for the auxiliary field $A^{-a}$, substitute the solution into the action \p{action} and integrate over harmonics. Skipping the details (see Refs. \cite{FIS-nonlin,glasses} for pedagogical derivations and the proof that the  metric  thus derived is hyper-K\"ahler), we quote the result. The component bosonic Lagrangian can be presented in the form
     \be
       L^{\rm bos} \ =\ \frac 12 g_{jc, kb} \,   {\dot x}^{jc}  {\dot x}^{kb} \,,
        \ee
        where
        \be
        \lb{metr-HK}
        g_{jc, kb}   = e^{\u{ja}}_{jc}  \, e^{\u{kb}}_{kb} \, \varepsilon_{\u{jk}} \, \Omega_{\u{ab} }
       \ee
       In this expression, the ordinary indices are the symplectic world indices, while the underlined indices refer to the tangent space.
       The vielbeins $e^{\u{kb}}_{kb}$ are determined from the relation
        \be
        \lb{vielbein-def}
        M^{\u{b}}{}_a (\partial_{kb} f^{+a} ) \ =\ e^{\u{kb}}_{kb} u_{\u{k}}^+\, ,
         \ee
         where the {\it bridge} $M^{\u{b}}{}_a$ is determined as a solution to the equation
         \be
         \lb{eq-bridge}
         {\cal D}^{++}M^{\u{b}}{}_a  = \partial^{++} M^{\u{b}}{}_a + \Omega^{cb} \, \frac {\partial^2 {\cal L}^{+4}}{\partial f^{+a} \partial f^{+b}} M^{\u{b}}{}_c \ =\ 0 \, .
          \ee
          {\it We reiterate:}

          \begin{itemize}

          \item To determine the HK metric, given the prepotential ${\cal L}^{+4}$, we first find $f^{+a}$ as a solution to the equation \p{cons-comp}.
          \item Then we solve the equation \p{eq-bridge} for the bridge.
          In fact, this homogeneous equation has  infinitely
    many solutions  interrelated by the transformation (a kind of gauge transformation)
      \be
      \lb{tau-gauge}
       M^{\u{b}}{}_a \ \to \ R^{\u{b}}{}_{\u{c}}  \, M^{\u{c}}{}_a
        \ee
         with an arbitrary harmonic-independent $R^{\u{b}}{}_{\u{c}}$. And for an arbitrary $R$, the objects
         $e^{\u{kb}}_{kb}$ defined in \p{vielbein-def} are not strictly speaking the vielbeins: the metric is not expressed via
         $e^{\u{kb}}_{kb}$ according to \p{metr-HK}, but its expression is more complicated. It is possible, however, to choose the gauge such that the expression \p{metr-HK} is valid. Note that the ``connection"
         $\Omega^{cb} \,  {\partial^2 {\cal L}^{+4}}/{\partial f^{+a} \partial f^{+b}}$ that enters \p{eq-bridge} belongs to the algebra $sp(2n)$. It is then possible to choose the solution for $M^{\u{b}}{}_a$ that belongs  to the group
         $Sp(2n)$. One can show that in this case the relation \p{metr-HK} holds.
         \item With the bridge in hand, we find the vielbeins using \p{vielbein-def} and then the metric.
          \end{itemize}

This is a long way, but it works.

  \section{A metric from Taub-NUT family}
  \setcounter{equation}0
   \subsection{Taub-NUT metric}
   To derive the Taub-NUT metric, we set $n=1$ and choose
   \be
\lb{L+4TN}
{\cal L}^{+4} \ =\ -\frac 1 2\, (q^{+1})^2 (q^{+2})^2 \, .
\ee
Having performed all the operations outlined above, we obtain
 \be
 \lb{metrx-TN}
ds^2 = \frac{2 +  x^1 \!\cdot\! x^2}{1+ x^1 \!\cdot \! x^2} \,
(x^2 \!\cdot \! dx^1  +  x^1 \!\cdot \! dx^2)^2
   +\,4(1+  x^1 \!\cdot\! x^2) dx^1 \!\cdot\! dx^2\,, \qquad  \label{ds-TN1}
 \ee
where $  x^1 \!\cdot\! x^2 = x^{j1} x_{j2}$, etc.

To bring the metric \p{metrx-TN} to the form known in the literature, we introduce the variables
\be
\lb{def-3X}
X^1 = x^{21} x^{22} - x^{11} x^{12},\ \  X^2 &=& i(x^{11} x^{12} + x^{21} x^{22}), \ \
X^3 = x^{11} x^{22} + x^{21} x^{12}, \nn
\Psi  &=&  i\,\ln \frac{x^{22}}{x^{11}}.
\ee
Then \p{metrx-TN} coincides up to the extra irrelevant factor 2 with
\be
\lb{metrX-TN}
ds^2\ =\ V(r) \, d\vecg{X}d\vecg{X} +  V^{-1}(r)\,  \left(d\Psi+ \vecg{A}d\vecg{X}\right)^2 \,,
\ee
where $\Psi \in (0, 2\pi)$,
\begin{equation}\label{V-TN-12}
V(r) \ = \ \frac{1}{r} + 1
\end{equation}
($r = |\vec{X}|$) and
\be
\label{U-TN}
A^1 \ =\  \frac{X^2}{r\left(r+X^3 \right)}\, , \ \ \
A^2 \ = \  -\frac{X^1}{r\left(r+X^3 \right)}\, , \ \ \
A^3 \ =\ 0
\ee
is the vector potential of a magnetic monopole.
The metric \p{metrX-TN} has nontrivial physical applications. In particular, a similar metric, but with the inverse sign of the ``Coulomb  potential  term" $1/r$ in \p{V-TN-12} describes the dynamics of
two interacting BPS magnetic monopoles  at the distances much larger than their size. \cite{Gib-Man}.\footnote{Such metric involves a singularity  at $r=1$. Physically, this singularity smoothes out: the dynamics of BPS monopoles at small distances is described by a more complicated Atiyah-Hitchin metric \cite{AH}, which is smooth and goes over to the TN metric in the limit of large $r$. These details are irrelevant for us, but we have chosen to mention them to give a reader a proper perspective.}

\subsection{8-dimensional calculation}
We take $n=2$ and choose
 \be
 \lb{L+4TN}
 L^{+4} = -q^{+1} q^{+2} q^{+3}q^{+4}\,.
  \ee
  The equations \p{cons-comp} read

  \be
\partial^{++} f^{+1} \ =\ f^{+1} f^{+3} f^{+4}, \qquad \partial^{++} f^{+2} \ =\ - f^{+2} f^{+3} f^{+4},  \nn
\partial^{++} f^{+3} \ =\ f^{+3} f^{+1} f^{+2}, \qquad \partial^{++} f^{+4} \ =\ - f^{+4} f^{+1} f^{+2}.
 \ee

 Their solution is

 \be
 f^{+1} \ =\ x^{+1} e^{\cal K}, \quad  f^{+2} \ =\ x^{+2} e^{- {\cal K}},
 \quad  f^{+3} \ =\ x^{+3} e^{\cal J}, \quad  f^{+4} \ =\ x^{+4} e^{- {\cal J}},
  \ee
  where
  \be
  {\cal J} \ =\ \frac 12 (x^{+1} x^{-2} + x^{-1} x^{+2} ), \qquad
{\cal K} \ =\ \frac 12 (x^{+3} x^{-4} + x^{-3} x^{+4} ).
    \ee
    Our next task is to solve the equation \p{eq-bridge}. As was mentioned above, this equation has many solutions, and to ensure \p{metr-HK}, we have to pick up a solution that belongs to $Sp(4)$. However, this requirement does not rigidly fix the solution yet: one can still multiply it by a constant $Sp(4)$ matrix. It is convenient to choose the solution which reduces to the unit matrix when the nonlinearity associated with ${\cal L}^{+4}$ is switched off.

    The bridge reads

     \be
   M^{\u{b}}{}_a  \ =\ \frac 1{\Delta^{1/4}} \left( \begin{array}{cccc}
     e^{-{\cal K}} & 0 & - x^{-1} x^{+4} e^{-{\cal J}} & - x^{-1} x^{+3} e^{\cal J} \\
     0 & e^{\cal K} & x^{-2} x^{+4} e^{-{\cal J}} & x^{-2} x^{+3} e^{\cal J} \\
     -x^{-3} x^{+2} e^{- {\cal K}} & - x^{-3} x^{+1} e^{\cal K} & e^{-{\cal J}} & 0 \\
     x^{-4} x^{+2} e^{- {\cal K}} & x^{-4} x^{+1} e^{{\cal K}} & 0 & e^{\cal J}
      \end{array}
      \right),
     \ee
    where the index $\u{b}$ marks the lines, the index $a$ marks the columns,
    \be
    \lb{Delta}
    \Delta = 1 - (x^1 \cdot x^2) (x^3 \cdot x^4),
    \ee
     and we used the notation $(a \cdot b) \equiv a^j b_j$.

     The derivatives of $f^{+a}$ that enter the equation  \p{vielbein-def} are:

     \be
     &&\partial_{k1} f^{+1} \ =\  u^+_k e^{\cal K}, \qquad
     \partial_{k2} f^{+2} \ = \ u^+_k e^{-\cal K}, \qquad   \partial_{k1} f^{+2} =  \partial_{k2} f^{+1} \ =\  0\,, \nn
      &&\partial_{k3} f^{+3} \ =  \ u^+_k e^{\cal J}, \qquad   \partial_{k4} f^{+4} \ = \ u^+_k e^{-\cal J}, \qquad
       \partial_{k3} f^{+4} = \partial_{k4} f^{+3} \ =\  0\,, \nn
     &&\partial_{k1} f^{+3} = \frac 12 x^{+3} (u^+_k x^{-2} + u^-_k x^{+2}) e^{\cal J},
     \qquad \partial_{k1} f^{+4} = - \frac 12 x^{+4} (u^+_k x^{-2} + u^-_k x^{+2}) e^{-\cal J}, \nn
      &&\partial_{k2} f^{+3} = \frac 12 x^{+3} (u^+_k x^{-1} + u^-_k x^{+1}) e^{\cal J}, \qquad \partial_{k2} f^{+4} = -\frac 12 x^{+4} (u^+_k x^{-1} + u^-_k x^{+1}) e^{-\cal J}, \nn
      &&\partial_{k3} f^{+1} = \frac 12 x^{+1} (u^+_k x^{-4} + u^-_k x^{+4}) e^{\cal K}, \qquad \partial_{k3} f^{+2} = -\frac 12 x^{+2} (u^+_k x^{-4} + u^-_k x^{+4}) e^{-\cal K}, \nn
      &&\partial_{k4} f^{+1} = \frac 12 x^{+1} (u^+_k x^{-3} + u^-_k x^{+3}) e^{\cal K},
  \qquad
   \partial_{k4} f^{+2} = - \frac 12 x^{+2} (u^+_k x^{-3} + u^-_k x^{+3}) e^{-\cal K}\,.
   \ee

  This gives the following vielbeins:

   \be
   e^{\u{k1}}_{k1} &=&  e^{\u{k2}}_{k2} \ = \  e^{\u{k3}}_{k3} \ = \ e^{\u{k4}}_{k4} \ = \ \Delta^{-1/4} \delta^{\u{k}}_k\,, \nn
   e^{\u{k1}}_{k2}&=&  0, \quad  e^{\u{k1}}_{k3} = \frac {\Delta^{-1/4}} 2 [\delta^{\u{k}}_k (x^1 \cdot x^4) - x^{\u{k}4} x^1_k],  \quad  e^{\u{k1}}_{k4} = \frac  {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^1 \cdot x^3) - x^{\u{k}3} x^1_k], \nn
    e^{\u{k2}}_{k1} &=& 0, \quad  e^{\u{k}2}_{k3} = \frac {\Delta^{-1/4}} 2 [\delta^{\u{k}}_k (x^4 \cdot x^2)  + x^{\u{k}4} x^2_k],  \quad  \ e^{\u{k2}}_{k4} = \frac {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^3 \cdot x^2) + x^{\u{k}3} x^2_k], \nn
     e^{\u{k3}}_{k4} &=& 0, \quad   e^{\u{k3}}_{k1} = \frac {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^3 \cdot x^2) - x^{\u{k}2} x^3_k],  \quad  e^{\u{k3}}_{k2} = \frac {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^3 \cdot x^1) -  x^{\u{k}1} x^3_k], \nn
     e^{\u{k4}}_{k3} &=& 0, \quad  e^{\u{k4}}_{k1} = \frac {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^2 \cdot x^4) + x^{\u{k}2} x^4_k],  \quad  e^{\u{k4}}_{k2} = \frac {\Delta^{-1/4}}2 [\delta^{\u{k}}_k (x^1 \cdot x^4)] +  x^{\u{k}1} x^4_k]\,.
   \ee

  The components of the metric are

   \be
   \lb{metr}
  \Delta^{1/2} g_{j1, k1} &=& \frac  14 [(x^2 \cdot x^4) (23)_{jk} - \{3 \leftrightarrow 4\} ], \qquad
     \Delta^{1/2}g_{j2, k2} \ = \ \frac 1 4 [(x^1 \cdot x^4) (13)_{jk} - \{3 \leftrightarrow 4\} ], \nn
     \Delta^{1/2}g_{j3, k3} &=& \frac 1 4 [(x^1 \cdot x^4) (24)_{jk} - \{1 \leftrightarrow 2\} ], \qquad
     \Delta^{1/2}g_{j4, k4} \ =\ \frac  14 [(x^1 \cdot x^3) (23)_{jk} - \{1 \leftrightarrow 2\} ],\nn
     \Delta^{1/2}g_{j1, k3}
     &=&  [\varepsilon_{jk} (x^4 \cdot x^2) + x^4_j x^2_k], \qquad
     \Delta^{1/2} g_{j1, k4} \ = \ [\varepsilon_{jk} (x^3 \cdot x^2) + x^3_j x^2_k], \nn
      \Delta^{1/2}g_{j2, k3} &=& [\varepsilon_{jk} (x^4 \cdot x^1) + x^4_j x^1_k], \qquad
      \Delta^{1/2}g_{j2, k4} \ =\ [\varepsilon_{jk} (x^3 \cdot x^1) + x^3_j x^1_k], \nn
      \Delta^{1/2}g_{j1, k2} &=&   \varepsilon_{jk} \left[ 1 + \frac 14 (x^2 \cdot x^4) (x^1 \cdot x^3)
     - \frac 14 (x^2 \cdot x^3) (x^1 \cdot x^4) \right]  \nn
    && + \ \frac 14 \left[  (x^2 \cdot x^4) x^1_j x^3_k + (x^1 \cdot x^4) x^3_j x^2_k
     - (x^1 \cdot x^3) x^4_j x^2_k - (x^2 \cdot x^3) x^1_j x^4_k\right]  \nn
    && + \ \frac 14 (x^1 \cdot x^2) (x^3_j x^4_k - x^4_j x^3_k), \nn
      \Delta^{1/2}g_{j3, k4} &=&   \varepsilon_{jk} \left[ 1 + \frac 14 (x^2 \cdot x^4) (x^1 \cdot x^3)
     - \frac 14 (x^2 \cdot x^3) (x^1 \cdot x^4) \right]  \nn
     &&+ \ \frac 14 \left[  (x^1 \cdot x^4) x^3_j x^2_k + (x^1 \cdot x^3) x^2_j x^4_k
     - (x^2 \cdot x^4) x^3_j x^1_k - (x^2 \cdot x^3) x^1_j x^4_k\right]  \nn
     &&+ \ \frac 14 (x^3 \cdot x^4) (x^1_j x^2_k - x^2_j x^1_k),
     \ee
where $(23)_{jk} = x^2_j x^3_k + x^3_j x^2_k$, etc.

\subsection{Discussion}

If $x^a \gg 1$, the metric simplifies, acquiring a block-diagonal form. The metric components behave then as $g_{MN} \sim x^2$. The invariant volume
$\int \! \sqrt{g} \, d^8x$ diverges.  The metric has thus similar properties to the Taub-NUT metric \p{metrx-TN} and represents an 8-dimensional generalization of the latter.

Three 8-dimensional generalizations of the Taub-NUT metric are known. They are related to the three classical simple Lie algebras of rank 2: $su(3), \ sp(4) = so(5)$, and $g_2$. The explicit expression for the
$su(3)$  metric is \cite{N-monopoles}:
  \be
  \lb{su3}
  ds^2[su(3)] \ =\ \sum_{m,l = 1,2,3} A_{ml} \, d \vecg{X}_m d \vecg{X}_l \quad + \ {\rm phase\ part}\,,
   \ee
   where $\vecg{X}_m$ are 3-dimensional vectors satisfying the constraint $\vecg{X}_1 + \vecg{X}_2 + \vecg{X}_3 = 0$,
   \be
   \lb{Amm}
   A_{mm} \ =\  1 + C \sum_{l \neq m} \frac 1{|\vecg{X}_m - \vecg{X}_l |} \qquad \qquad ({\rm no\ summation \ over\ } m)
   \ee
   and
   \be
   \lb{Aml}
  A_{m \neq l} \ =\   - \frac C{|\vecg{X}_m - \vecg{X}_l|}\, ,
   \ee
  where $C$ is a real constant.

   This metric (with negative $C$ !) describes the dynamics of three well-separated BPS monopoles. Then $\vecg{X}_m$ are their positions.

  The explicit expressions for the $so(5)$ and $g_2$  metrics were found in \cite{Selivanov}. They are:
    \be
  \lb{so5}
  ds^2[so(5)] \ =\ \sum_{m=1,2}  d \vecg{X}_m^2  +  C\left[\sum_{\pm}  \frac
  {(d\vecg{X}_1 \pm d\vecg{X}_2)^2}{|\vecg{X}_1 \pm \vecg{X}_2|} + \sum_{m = 1,2} \frac {d\vecg{X}_m^2}{X_m}\right] + \ {\rm phase\ part}
   \ee
   and
    \be
  \lb{g2}
  ds^2[g_2] \ =\ \sum_{m=1}^3 d \vecg{X}_m^2  +  C\left[\sum_{l<m\leq3}  \frac
  {(d\vecg{X}_m - d\vecg{X}_l)^2}{|\vecg{X}_m -  \vecg{X}_l|} + 3\sum_{m=1,2,3} \frac {d\vecg{X}_m^2}{X_m}\right] + \ {\rm phase\ part}
   \ee
 with the constraint $\vecg{X}_1 + \vecg{X}_2 + \vecg{X}_3 = 0$.
  The reader has recognized in the structures entering the right-hand sides of Eqs. \p{Amm} - \p{g2} the roots of the corresponding algebras.

The metric \p{so5} can be obtained from the $su(4)$ metric describing the dynamics of four BPS monopoles by the hamiltonian hyper-K\"ahler reduction in the spirit of \cite{Hitchin,Rychenkova}. The metric \p{g2} can be obtained by a reduction of the  metric describing the dynamics of seven monopoles. (7 is the lowest dimension of a unitary group where $G_2$ can be embedded.)

A natural conjecture is that the newly derived metric \p{metr} is equivalent to one of these known hyper-K\"ahler metrics. If so, it is clear with  which one. The metric \p{metr} has the $Z_2$ isometry corresponding to the simultaneous interchange\footnote{It also  has  the continuous $SU(2)$ isometry associated with the rotations of the spinor indices $j,k$. This isometry, associated with the rotations of $\vecg{X}_m$, is present also in \p{so5} [and in \p{su3}, \p{g2}].} $x_1 \leftrightarrow x_3\, , x_2 \leftrightarrow x_4$. Also the metric \p{so5} has the $Z_2$ isometry associated with the interchange $\vecg{X}_1 \leftrightarrow \vecg{X}_2$. On the other hand, the metrics \p{su3} and \p{g2} have both the $S_3$ isometry  associated with the interchange of $\vecg{X}_{m=1,2,3}$, which the metric \p{metr} does not possess. (Obviously, these isometries are related to the Weyl groups of the corresponding algebras.)

The reader may ask at this point: how come the isometry of the metric \p{metr} is only $Z_2$, while the harmonic prepotential \p{L+4TN}, from which this metric was derived was invariant under $S_4$ ? This apparent paradox is resolved by noting that we have chosen the particular convention \p{Omega} for the symplectic matrix $\Omega^{ab}$. The transformation $(12) \leftrightarrow (34)$ is
the only discrete transformation leaving this matrix invariant.

Thus, we guess that the metric \p{metr} can be reduced to \p{so5} by a coordinate transformation expressing $\vecg{X}_m$ as a quadratic function of $x$---an analog of the transformation \p{def-3X}. It would be interesting to find this transformation explicitly and confirm this guess.

Given that the prepotential \p{L+4TN} gives the metric \p{so5}, a natural question is what prepotentials give the metrics \p{su3} and \p{g2}. Unfortunately, we cannot now answer this question. The procedure described in Sect. 2 is ``asymmetric": we can obtain the metric from the prepotential, but we  know no regular way to derive the prepotential from the metric. On the other hand, an exact mathematical proof of the fact that {\it any} HK metric can be derived from {\it some} prepotential ${\cal L}^{+4}(q^{+a})$ has been constructed \cite{L+4-HK-teor}.

  \section*{Appendix: Symplectic notation}
  \setcounter{equation}0

  \renewcommand\theequation{A.\arabic{equation}}

 We trade the vector index $M$ for a pair $(ja)$, where $j = 1,2$ and $a = 1, \ldots, 2n$.
  The indices are raised and lowered according to $X^i = \varepsilon^{ij}X_j, Y^a = \Omega^{ab} Y_b$,
 where $\varepsilon^{jk} = -\varepsilon_{jk}$ with the convention $\varepsilon_{12} = 1$ and
$\Omega^{ab}$ is the symplectic matrix, which we choose in the form \p{Omega}.

 We introduce $4n$ rectangular matrices $\Sigma_M$,
\be
\label{Sigma}
(\Sigma_{1,2,3,4})^{ja} = \left( \sigma^\dagger_\mu\,, \, 0\,, \, \ldots \,,\, 0  \right)^{ja}, \
(\Sigma_{5,6,7,8})^{ja}  = \left( 0\,,\, \sigma^\dagger_\mu\,, \, 0\,, \, \ldots \,,\, 0  \right)^{ja}, {\rm etc.}
\ee
with\footnote{And $(\sigma_\mu)_{aj} \  = \ \{(\vecg{\sigma})_{aj}, i\delta_{aj} \}$. Note the identities
$$ \sigma_\mu \sigma_\nu^\dagger + \sigma_\nu \sigma_\mu^\dagger = 2 \delta_{\mu\nu}\,. $$}
\be
\lb{sigma-mu}
(\sigma^\dagger_\mu)^{ja} \  = \ \{(\vecg{\sigma})^{ja}, -i\delta^{ja} \}     \, .
\ee
In \p{sigma-mu}, $a = 1,2$ and  $(\vecg{\sigma})^{ja}$ as well as $(\vecg{\sigma})_{aj}$ are the  {\it Pauli matrices}.

Then, for any tensor, we establish the correspondence
 \be
 \lb{sympl-vector}
T^{\ldots {ja} \ldots}  =  \frac i{\sqrt{2}} \, (\Sigma_M)^{ja}\, T^{\ldots M \ldots} \,, \quad
T^{\ldots M \ldots}  =  \frac i{\sqrt{2}} \, (\Sigma_M)_{ja}\, T^{\ldots ja \ldots}, \quad
 \ee
 where $(\Sigma_M)_{ja} = \varepsilon_{jk} \Omega_{ab} (\Sigma_M)^{kb}$,
  and
the dots stand for all other indices. Note that, for a real vector $V^M$, the components $V^{ja}$ obey the  {\it pseudoreality} condition
\be
\lb{pseudoreal-V}
\overline{V^{ja}} \ =\ \varepsilon_{jk} \Omega_{ab}  V^{kb} \equiv V_{ja} \, .
\ee

In these terms, the flat metric is expressed as
\be
\lb{metr-simpl}
g^{ja,\, kb} \ =\  -\frac 12\, (\Sigma_M)^{ja}  (\Sigma_M)^{kb}  =
\varepsilon^{jk} \Omega^{ab}  \, , \nn
g_{ja,\, kb} \ =\   \varepsilon_{jk} \Omega_{ab} \ = \ g^{ja,\, kb}\, .
\ee

The quaternionic triple of the tangent space complex structures can be chosen as
\be
\lb{Ip}
(I^p)_{ja, kb} \ =\
-i(\sigma^p)_{jk} \, \Omega_{ab}\, ,
 \ee
 where
 \be
\lb{symm-jk}
(\sigma^p)_{jk} \ =\ (\sigma^p)_{kj} \ =\ \varepsilon_{kl}
(\sigma^p)_{j}{}^{l}
  \ee
[$(\sigma^p)_{j}{}^{l}$ are the ordinary Pauli matrices].

\vspace{1mm}

config: arxiv


TITLE: What is a Cheeger deformation?
QUESTION [8 upvotes]: I'm currently at a Differential Geometry meeting and there is a mini-course on positively curved Riemannian manifolds. There, we were told that a technique to construct such manifolds is a Cheeger deformation, which (if I understood correctly) is a generalization of a one-parameter family of surfaces of revolution given by $\frac{f}{\lambda f + 1}$, where $f$ is the curve that generates a surface of revolution in Euclidean space and $\lambda$ is a positive parameter that varies over $[0,\infty)$. Can anyone tell me what is the concrete definition of a Cheeger deformation and how are they used to construct manifolds of positive (or non-negative, perhaps; I don't remember) curvature? Thanks a lot.

REPLY [15 votes]: Cheeger deformations were first introduced in Jeff Cheeger's 1973 paper Some examples of manifolds of nonnegative curvature, and were inspired by the Berger sphere construction. The basic idea is to consider a Lie group acting by isometries on a Riemannian manifold, and then shrink the lengths of vectors tangent to the orbits of the action while keeping those perpendicular to the orbits fixed. Here is a formal description:

Let $(M,g_M)$ be a Riemannian manifold with $G$, a closed subgroup of the isometry group of $M$, acting on the left. Choose any bi-invariant metric $g_G$ on $G$, and consider the one-parameter family of product metrics $\{l^2g_G + g_M\}_{l>0}$ on the product manifold $G\times M$. Then $G$ acts on $G \times M$ via the map
$$y\cdot(x,p) = (xy^{-1}, y\cdot p).$$
This action is free, so taking the quotient gives a manifold $(G\times M)/G$, and it is diffeomorphic to $M$. Thus by imposing that the quotient map
$$q_l:(G \times M, l^2g_G + g_M)\to (G\times M)/G\cong M$$
be a Riemannian submersion, we obtain a one-parameter family of metrics $g_l$ on $M$. The family of Riemannian manifolds $\{(M,g_l)\}_{l>0}$ is called a Cheeger deformation of the original manifold $(M,g_M)$.

As $l\to \infty$, the Cheeger deformed metrics $g_l$ converge to the original metric $g_M$, and as $l\to 0$, the manifolds $(M,g_l)$ converge in the Gromov-Hausdorff metric to the quotient space $M/G$. Moreover, if the original manifold $(M,g_M)$ is one of nonnegative (resp. positive) sectional curvature, then the Cheeger deformed manifolds $(M,g_l)$ will have nonnegative (resp. positive) sectional curvature for all values of $l$.
The survey by Wolfgang Ziller mentioned by @Dan in the comments is a great reference for general information about these deformations and how they affect curvature. See my blog post for an animation of the Cheeger deformed plane under the circle action given by rotation about the origin.


TITLE: What are the open sets in the quotient space formed when the endpoints of a closed interval are "glued together"?
QUESTION [0 upvotes]: I understand the definition of the quotient topology, but can't intuitively understand how the open sets are formed.

REPLY [2 votes]: The gluing process is made precise by following this procedure, which will be familiar to you, but is worth writing out in detail, to see what the open sets in the quotient topology are:
$1).\ $ if $[a,b]$ is a closed interval, define an equivalence relation $\sim$ by declaring the classes to be: the singleton sets $\{\{x\}:x\in (a,b)\}$ and the set $\{a,b\}.$ All this means is that we have defined a new set $[a,b]/\sim$ which consists of points in the interior of the interval $[a,b]$ and the "point" $\{a,b\}$.
$2).\ $ To topologize  $[a,b]/\sim$, we define a function $q:[a,b]\to [a,b]/\sim$ by sending each $x\in [a,b]$ to its equivalence class. So, for example, if $a<x<b,\ q$ sends $x$ to $\{x\}$, which we may as well just take to be $x$ itself.  But if $x=a$ or $x=b,$ then $q$ sends $x$ to the class $\{a,b\}$ which is a new "point", which we call $[a]\ (\text{or}\ [b])$.
Now, we define a set $V$ in $[a,b]/\sim$  to be open if and only if $q^{-1}(V)$ is open in $[a,b].$
To see what the open sets are, first note that if $V$ is an open interval $(x,y)$ with $x,y\neq a$ or $b$, then $q^{-1}(V)=V$ itself, which is open in $[a,b]$, and so $V$ is open in $[a,b]/\sim$. This is no surprise, since $q$ sends all interior points of $[a,b]$ to themselves (identifying $x$ with $\{x\}$). Now check some other possibilities for $V$:
If $V=[a]$, with then $q^{-1}(V)$ is $\{a,b\}$ which is not open in $[a,b]$ and so the point $[a]$ is not open in $[a,b]/\sim$.
If $V=[a]\cup (a,x),\ x<b$, then $q^{-1}(V)=[a,x)\cup\{b\}$ which is not open in $[a,b]$ so $V$ is not open in $[a,b]/\sim$.
If $a<x<y<b$ and $V=[a]\cup (a,x)\cup (y,b)$, then $q^{-1}(V)=[a,x)\cup (y,b]$, which is open in $[a,b]$ so $V$ is open in $[a,b]/\sim$.
Let $a<x<y<b$. It should be becoming clear that the basic open sets in the quotient are those sets $V$ for which $q^{-1}(V)$ is of the form $(x,y)$ or $[a,x)\cup (y,b]$. And since $a$ and $b$ are identified in the quotient, this latter set may be considered to be a "gluing" of $[a,x)$ and $(y,b]$ into a "single" open set, which after all, is what the quotient was supposed to do.


TITLE: Why the minimum of this multivariate polynomial is not a critical point?
QUESTION [2 upvotes]: I have this multivariate polynomial function
$$\begin{align*} f(L,x,y,z) &= L^2 + (1 - L x)^2 + (1 - L y)^2 + (1 - L z)^2 \\
&+ (2 - L x y)^2 + (2 - L x z)^2 + (2 - L y z)^2 + (3 - L x y z)^2
\end{align*}$$
To compute its minimum an idea is to computes its critical points and then verify which of them minimizes $f$. This is a differentiable function, so all maxima and minima of $f$ are critical points, right?
Solving the equation $\nabla f(L,x,y,z) = (0,0,0,0)$ we can find its critical points. Supposing $L \neq 0$, we have the following polynomial system.
$$ \begin{cases}
  \begin{align*} X &+ Y + Z + 2 X Y + 2 X Z + 2 Y Z + 3 X Y Z \\
&-L(1 + X^2 + Y^2 + Z^2 + X^2 Y^2 + X^2 Z^2 + Y^2 Z^2 + X^2 Y^2 Z^2) = 0
\end{align*}\\
  -1 + Y - Y^2 + Z + 2 Y Z - Z^2 - Y^2 Z^2 = 0\\
  -1 + X - X^2 + Z + 2 X Z - Z^2 - X^2 Z^2 = 0\\
  -1 + X - X^2 + Y + 2 X Y - Y^2 - X^2 Y^2 = 0\\
 \end{cases}$$
I tried to solve this system using Mathematica and PHCPack. Both programs solved the system without a problem, but both missed the actually minimum. I had to use a minimizer routine in Mathematica to get the minimum $(L,x,y,z) = \left( \frac{1+\sqrt{17}}{9}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4}, \frac{3+\sqrt{17}}{4} \right)$. This point minimizes $f$ but is not a solution of the system
above.
My question is, why this minimum is not a critical point? How is that possible?

REPLY [2 votes]: Reduce[Grad[L^2+(1-L x)^2+(1-L y)^2+(2-L x y)^2+(1-L z)^2+(2-L x z)^2+(2-L y z)^2+
(3-L x y z)^2,{L,x,y,z}]=={0,0,0,0}&&L!=0,{L,x,y,z}]


TITLE: Method of Proof
QUESTION [1 upvotes]: Consider a theorem of the form: (Something) iff $\exists$ a continuous function $f: X \rightarrow Y$ s.t. (something about the function).
I'm unsure how one would go about proving the $(\rightarrow)$ direction.
So we suppose that (Something) is true, and we WTS that there exists a continuous ...
But in terms of showing the part on the right, do we let $f$ be a continuous function and then show the s.t. portion? Or do we let $f$ be a function s.t. (something about the function) and then show that $f$ is continuous? Or am I only allowed to let $f: X \rightarrow Y$ and then have to show $f$ is continuous and the s.t. (something about the function) follows as well?
Thanks!

REPLY [1 votes]: Do we let $f$ be a continuous function and then show the s.t. portion?

No, because then you would show that every continuous function has that portion, which is not always true.

Or do we let $f$ be a function s.t. (something about the function) and then show that $f$ is continuous?

No, because then you would show that every function with that portion is continuous, which is also not always true.

Or am I only allowed to let $f:X\to Y$ and then have to show $f$ is continuous and the s.t. (something about the function) follows as well?

No, because then you would prove that every function is continuous and has that portion. Which is not true.

What you need to do: assume (something), then construct a continuous function with that portion. So you really need to make a continuous function yourself and prove that it has that portion (all based on the assumption of something).


TITLE: Integral of a piecewise function
QUESTION [3 upvotes]: Check whether the integral $\int_{-1}^2{f(x)[2x+1]}dx$ exists where
  $$f(x)=
\begin{cases}
x+2, &\text{$x<0$}\\
1, & \text{$x=0$}\\
xe^{x^2},& \text{$x>0$}
\end{cases}$$
  and $[x]$ denotes the greatest integer function. If it exists then calculate the integral.

I'm clueless about how to do the whole proof by definition stuff. Is there a way to prove that this function is integrable without any epsilon/delta and partition proofs?

REPLY [3 votes]: Because of the fact that the discontinuities of $f(x)$ are of measure zero, the function is Riemann integrable. A proof of this exists here. This means we can validly take the integral of $f(x)$ on either side of the discontinuity and sum those:
$$\int_{-1}^{2} f(x)[2x+1]dx = \int_{-1}^{0}(x+2)[2x+1]dx + \int_{0}^{2}xe^{x^2}[2x+1]dx$$
Next, because $[2x+1]$ is the same value on for any x in the interval $[n, n+\frac{1}{2})$ where $n \in \mathbb{Z}$ and these new discontinuities are also of measure zero, we can seperate this into a larger sum with each value of $[2x+1]$ written out.
$$-1 \cdot \int_{-1}^{-\frac{1}{2}} (x+2)dx + 0 \cdot \int_{-\frac{1}{2}}^{0} (x+2)dx + 1 \cdot \int_{0}^{\frac{1}{2}} xe^{x^2}dx + 2 \cdot \int_{\frac{1}{2}}^{1} xe^{x^2}dx + 3 \cdot \int_{1}^{\frac{3}{2}} xe^{x^2}dx + 4 \cdot \int_{\frac{3}{2}}^{2} xe^{x^2}dx$$
Now we remove the term that is multiplied by $0$ and find $\int_{a}^{b} xe^{x^2}dx$ by setting $u=x^2$ and finding that $du=2xdx$.
$$-\int_{-1}^{-\frac{1}{2}} (x+2)dx + \frac{1}{2}\int_{0}^{\frac{1}{4}} e^{u}du + \int_{\frac{1}{4}}^{1} e^{u}du + \frac{3}{2}\int_{1}^{\frac{9}{4}} e^{u}du + 2\int_{\frac{9}{4}}^{4} e^{u}du$$
Finally we take the antiderivatives and evaluate.
$$-\left[\frac{x^2}{2}+2x\right]_{-1}^{-\frac{1}{2}}+\frac{1}{2}[e^u]_{0}^{1/4}+[e^u]_{1/4}^{1}+\frac{3}{2}[e^u]_{1}^{9/4}+2[e^u]_{9/4}^4$$
$$=-\left(\frac{1}{8}-1-\frac{1}{2}+2\right)+\frac{e^{1/4}-1}{2}+\left(e-e^{1/4}\right)+\frac{3e^{9/4}-3e}{2}+\left(2e^4-2e^{9/4}\right)$$
$$=-\frac{9}{8}-\frac{e^{1/4}}{2}-\frac{e}{2}-\frac{e^{9/4}}{2}+2e^4$$
$$=\frac{16e^4-4e^{9/4}-4e-4e^{1/4}-9}{8} \approx 101.326$$


TITLE: A ship captain sails 500 miles due south and then 1,200 miles due east
QUESTION [0 upvotes]: What is the minimum number of miles the captain must sail to return to his original position?
I got $\sqrt{500^2+1200^2}=1300$
The practice test I'm taking says the answer can't be determined which I'm $99.9\%$ sure is wrong but just want to double check.

REPLY [0 votes]: I think your answer is right. If this problem is given to many people 99.9% have the same answer. And if in practice set answer is not same as you calculated. Its exception.


TITLE: MacLane’s coherence theorem and free monoidal categories
QUESTION [2 upvotes]: Original question
The nLab gives one formulation of the coherence theorem for monoidal categories:

Every diagram in a free monoidal category made up of associators and unitors commutes.

It seems to me in practice this coherence theorem is often used as follows:
One has a diagram on a set of objects $S$ of a specific monoidal category $(C,\otimes,I)$ made up from the associators and unitors of $C$. One then considers this diagram "in the free monoidal category on $S$." Subsequently, one argues that since the diagram in the free monoidal category commutes, so does the original diagram.
How do you make this sketch of an argument precise? I tried using the universal property of „the free monoidal category“ – to no avail.
Edit:
In his answer Jonas suggested that any monoidal category is the quotient of a free monoidal category. Is this true? In the context of group theory this is easily proved using the universal property of the free group and the first isomorphism theorem for groups. Does an analogue of the isomorphism theorem hold for (monoidal) categories?

REPLY [4 votes]: The free monoidal category $\mathbb M A$ on a set $A$ has the following universal property. For every monoidal category $\mathbf C$ and function $f \colon A \to |\mathbf C|$, there is a unique monoidal functor $\tilde f \colon \mathbb M A \to \mathbf C$ such that $\tilde f \circ \eta_A = f$, where $\eta_A \colon A \to \mathbb M A$ is the inclusion (concretely, $\mathbb M A$ has as its objects lists of objects in $A$, and $\eta_A$ sends $a \mapsto [a]$).
A diagram in the free monoidal category on $A$ is simply a pair of morphisms $\ell, r \colon x \to y$ in $\mathbb M A$. Conceptually, these are built inductively from the structural isomorphisms $\alpha$, $\lambda$, and $\rho$.
A formal diagram in a monoidal category $\mathbf A$ is simply a diagram in $\mathbb M(|\mathbf A|)$. We have a canonical functor $\mathbb M(|\mathbf A|) \to \mathbf A$ given by the universal property of the free monoidal category. Therefore, a formal diagram in $\mathbf A$ is sent to a pair of morphisms in $\mathbf A$. Since the morphisms in the free monoidal category were constructed inductively from structural isomorphisms, so will be the morphisms in $\mathbf A$ in the image of the functor.
However, the coherence theorem says that necessarily that for any diagram $\ell, r$ in the free monoidal category, we have $\ell = r$. Since functors preserve equality of morphisms, the functor $\mathbb M(|\mathbf A|) \to \mathbf A$ sends any diagram to an equal pair of morphisms in $\mathbf A$. Therefore any formal diagram in $\mathbf A$ commutes.


TITLE: How to prove that $E(X\mid\sigma\,(\mathcal{G},\mathcal{M}))=E(X\mid\mathcal{M})$ if $\mathcal{G}$ is independent of $X$ and $\mathcal{M}$?
QUESTION [3 upvotes]: I want to prove the following:

Let $X\in L^1(\Omega,\mathcal{F},\mathbb{P})$ be a random variable and suppose $\mathcal{G,M}\subset\mathcal{F}$ are sub–$\sigma$–algebras such that $\mathcal{G}$ is independent of $X$ and $\mathcal{M}$. Prove that $$\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))=\mathbb{E}(X\mid\mathcal{M}).$$

Following hints from my textbook, I have proved that for all $G\in\mathcal{G}$ and $M\in\mathcal{M}$ $$\int_{G\cap M}\mathbb{E}(X\mid\mathcal{M})\,\text{d}\mathbb{P}=\int_{G\cap M}X\,\text{d}\mathbb{P}.$$
However, there's number of issues I do not understand and I hope you could clarify them to me.
I know that $\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))$ is the only r.v. $\eta$ such that
(i) $\eta$ is $\sigma\,(\mathcal{G},\mathcal{M})$–measurable,
(ii) $\forall_{A\in\sigma\,(\mathcal{G},\mathcal{M})} \int_A\eta=\int_AX.$
By showing that RHS of the equality in the problem satisfies (i) and (ii), I'm done. However, my questions are:

$1)$ Why is $\mathbb{E}(X\mid\mathcal{M})$ a $\sigma\,(\mathcal{G},\mathcal{M})$–measurable r.v.?
  $2)$ So far, I proved (ii) for $A$'s of the form $A_1\cap A_2$ where $A_1\in\mathcal{G}$, $A_2\in\mathcal{M}$. Does the claim follows from that? If so, why?

Thanks for help!

Edit: Ad question 2): I've just found that it is sufficient to verify (ii) for every set $A$ in some $\pi$–system that contains $\Omega$ and generates $\sigma\,(\mathcal{G},\mathcal{M})$. Is it true that $\{G\cap M\mid G\in\mathcal{G}, M\in\mathcal{M}\}$ is a generator of $\sigma\,(\mathcal{G},\mathcal{M})$? I will strongly appreciate any stories that may possibly improve my understanding of $\sigma$–algebras.

REPLY [2 votes]: Because $E[X\mid \cal M]$ is by definition $\cal M$-measurable and $\cal M\subset \sigma(\cal G,\cal M)$.
The idea which consists to use $\pi$-system is good. Let $\mathcal{P}:=\{G\cap M,G\in\mathcal{G},M\in\cal M\}$. Then $\cal P\supset \cal M,\cal G$ as $\Omega\in\cal M\cap\cal G$. If $G\in \cal G$ and $M\in\cal M$, then $G\in\sigma(\cal G,\cal M)$ and $M\in\sigma(\cal G,\cal M)$, and their intersection still  will be in $\sigma(\cal G,\cal M)$, proving that $\sigma(\cal P)=\sigma(\cal G,\cal M)$.

Now we conclude showing that the collection of subsets $S$ of $\Omega$ satisfying
$$\int_SE[X\mid \mathcal M]\mathrm{d}\Bbb P=\int_SX\mathrm{d}\Bbb P$$
is a $\lambda$-system.


TITLE: Can we create a potential difference across the ends of a metallic rod just by accelerating it?
QUESTION [1 upvotes]: If we take a metallic rod and accelerate it then what would happen?
My view is that, as a metal has a sea of free electrons flowing about these free electrons will experience a pseudo force and will somewhat get displaced. This can create a Potential Difference in the rod as the positively charged kernels are fixed in the metallic lattice and hence they wont get displaced.
So a net displacement of negative charge and hence a Potential difference across the ends of the rod?
Is my reasoning correct?

REPLY [1 votes]: No. They are already accelerating. What potential difference does is make electrons go in a particular direction,  which your external force cant. Even if all electrons does experience a pseudo force  their relative motion won't change between each electron.


TITLE: Gaussian Elimination: Find a matrix satisfying the conditions.
QUESTION [1 upvotes]: QUESTION
Let $\begin{array}{cc} x=[1&2&3&4]^T \end{array}$. Find a matrix $E = I - \alpha e_i e_4^T$
such that $Ex =$
$\begin{bmatrix}
1\\
2\\
0\\
4\end{bmatrix}$
Any help with this would be appreciated. The part I'm most confused about is the $e_i$ and $e_4^T$. What do these represent? I think I should be fine after that I'm just a little confused at the moment.

REPLY [1 votes]: Usually, the notation $e_j$, or sometimes $\hat{e}_j$, is used to denote the standard basis vectors of $\mathbb{R}^n$; $e_j$ is a vector that has a $1$ in the $j$-th component and $0$ in the remaining components. E.g.
$$ e_3 = \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right]$$
Do be aware, however, that many sources don't distinguish very well between whether a vector should appear as a row or a column when treated as a matrix. Don't be surprised to see some sources use $e_3$ to mean $\left[ \begin{matrix} 0 & 0 & 1 & 0 \end{matrix} \right]$ instead, or even switch back and forth between the two options!


TITLE: Does this Weighted Lp bound my Sobolev-esque Fourier Space?
QUESTION [0 upvotes]: I have the norms
$$
||f||_{L^{\infty,n}} = \sup_{x\in X}\left|f(x)\bigg(\frac{1}{1+||x||_2}\bigg)^n\right| \\
||f||_{F^{n,1}} = \int_{\mathbb{R}^d}(1+||\xi||)^n|\hat{f}(\xi)|d\xi
$$
with $X\subseteq \mathbb{R}^d$ for $d\in\mathbb{N}$, $\hat{f}$ the Fourier Transform of $f$ and $n\in\mathbb{N}$.
I would like to know whether

$||f||_{L^{\infty,n}} < \infty \implies ||f||_{F^{n,1}} < \infty$
$||f||_{F^{n,1}} < \infty \implies ||f||_{L^{\infty,n}} < \infty$

I have been able to show that 1. does not hold, since for $f(x)=\sin{(1/||x||)}$ it holds that $||f||_{L^{\infty,n}} = 1 < \infty$ but $\hat{f}$ does not exist.
So far I have not been able to show that 2. does or does not hold.
The things I think are useful for proving the statement or providing a counter example are

$\widehat{f\star g} = \hat{f}\hat{g}$
$\hat{\hat{f}}(x)=f(-x)$
If $f$ has absolutely continuous $k$-th partial derivates, then $|\hat{f}(\xi)|\leq \frac{C}{(1+||\xi||)^k}$
$||\hat{f}||_{L^\infty} = \sup_{\xi\in\mathbb{R}^d} |\hat{f}(\xi)|\leq ||f||_{L^1}$

I would like help with 2.. Could you help me prove the statement or provide a counter example?

REPLY [2 votes]: I believe I have a counterexample for (2).
One important thing about the Fourier transform is that it trades smoothness for rapid decrease and vice versa. Note that
$$\int (1+|\xi|)^n|\hat{f}(\xi))| \ d\xi < +\infty$$
implies that $\hat{f}(\xi)$ decreases more rapidly than $(1+|\xi|)^n$. This implies that $f$ will have some level of regularity, but not necessarily rapid decrease. On the other hand, we will have
$$\sup_X\left| f(x)\left( \frac{1}{(1 + |x|)^n} \right) \right| < +\infty$$
if $f$ grows no faster than $(1+|x|)^n$. However, a rapidly decreasing Fourier transform only gives smoothness, so it doesn't seem like (2) should hold. In particular, we just want a smooth function that grows faster than $(1+|x|)^n$.
Let's look at the simplest case: $d=1$. I don't see that you specify what $X$ must be, so I'll assume that you have some freedom in choosing and take $X=\mathbb{R}$. Based on the above, it seems intuitively that $f(x) = x^{n+1}$ will be a good candidate for a counterexample. Then,
$$\hat{f}(\xi) = 2\pi i^{n+1}\delta^{(n+1)}(\xi).$$
This tells us that
$$\int (1+|\xi|)^n |\hat{f}(\xi)| \ d\xi = 2\pi \int (1+|\xi|)^n\delta^{(n+1)}(\xi) \ d\xi < +\infty.$$
However,
$$\sup_\mathbb{R}\left| x^{n+1}\left( \frac{1}{(1+|x|)^n} \right) \right| = +\infty.$$
On the other hand, if $X$ is bounded I suspect that this might be true. Since $\|f\|_{F^{n,1}} < +\infty$ gives $f$ a certain amount of smoothness, it will be bounded on bounded sets.


TITLE: How intrinsic is the $d^d$ term in the running time for constructing $\varepsilon$-nets in range spaces of VC-dimension d?
QUESTION [8 upvotes]: An $\varepsilon$-net for a range space $(X,\mathcal{R})$ is a subset $N$ of $X$ such that $N\cap R$ is nonempty for all $R\in \mathcal{R}$ such that $|X\cap R| \ge \varepsilon |X|$.
Given a range space $(X,R)$ of VC-dimension $d$, an $\varepsilon$-net of size $O\left(\frac{d}{\varepsilon}\log\left(\frac{d}{\varepsilon}\right)\right)$ can be computed in time $O(d)^{3d}\left(\frac{1}{\varepsilon^2}\log\left(\frac{d}{\varepsilon}\right)\right)^d|X|$ (see [1], Thm 4.6).
To what extent is the $O(d)^{3d}$ term intrinsic to this problem?  Specifically, can it be improved to $2^{O(d)}$?  Are there lower bounds known?
A related question: are there general conditions on $(X,R)$ for which such an improvement is known to exist?
[1] Bernard Chazelle.  The Discrepancy Method.  2000.

REPLY [10 votes]: If you are ok with a randomized algorithm, then you can just randomly sample $O((d/\epsilon) \log (d/\epsilon \delta))$ points, and this will be an $\epsilon$-net with probability at least $1-\delta$.
For the deterministic algorithm, I believe the $d^{O(d)}$ term comes from the following reason.  You first partition your full set into subsets of size about $s = O((d/\epsilon) \log (d/\epsilon))$ where $s$ is the size of the $\epsilon$-net you hope to get in the end.  Then you merge two such subsets and reduce to half that size.  (Basically, this is repeated until one set is left.)  The reduction step essentially considers all interesting ranges of the merged set of size $2s$.  By VC-bounds there are about $O((2s)^d)$ ranges, and you need to "check" each of these to see if it has been hit.  The $s^d$ has a $d^{O(d)}$ in it.  The $d$ in the $s$ term is needed.  So to reduce this bound, using this deterministic framework, you would somehow need to implicitly "check" all ranges without explicitly enumerating them.   This seems difficult to do deterministically, but I am not sure I know an explicit lower bound - most work in this arena assumes $d$ is constant and abuses this heavily.


TITLE: let $A$ a nonempty subset of $ \mathbb{R} $ that is bounded above, then $ \bigcup A \in A $??
QUESTION [0 upvotes]: in this question $ \mathbb{R} $ is defined as set of Dedekind cuts on $ \mathbb{Q} $
let $ A $ a nonempty subset of $ \mathbb{R} $ that is bounded above, and $ \bigcup A:=\{x|\exists B \in A(x \in B)\} \in A $
I must proof that $ \bigcup A \in A $
Thanks in advance!!

REPLY [1 votes]: It's actually not correct: Let $B_n=\{ x\in \Bbb Q:x\lt 1-\frac 1n\}$ and $A=\{ B_n:n\in \Bbb Z^+\}$, then $\bigcup A=\{ x\in \Bbb Q:x\lt 1\} \notin A$.
Edit: Note that $\bigcup A=\mbox{sup}A,$ so really it suffices to take any subset of $\Bbb R$ which doesn't contain its supremum.


TITLE: Linear Algebra- inner product of two same vectors
QUESTION [0 upvotes]: Wonder that is the inner product of two same vectors a real number? (not necessarily the Euclidean inner product) I think it’s implied by the $\langle v,v \rangle$ greater or equal to $0$ and the conjugate symmetry stuff, yet found no verification online.
This way, the norm should also be real number since it’s defined to be the positive square root of the inner product of two same vectors?
Thanks!

REPLY [2 votes]: If you consider the Hermitian symmetry of inner product, then
$$ <v,v> = \overline{<v,v> } $$
(The bar represents complex conjugate), which then implies the inner product should be real.


TITLE: Division algorithm for multivariate polynomials?
QUESTION [18 upvotes]: We know that if $F$ is a field and $f(X)$ a non-zero polynomial in $F[X]$, then for every polynomial $g(X)$ we can find $q(X),r(X)$ such that
$$g(X)=f(X)\cdot q(X)+r(X)$$
with $r(X)$ the zero polynomial or $\deg r(X)<\deg f(X)$.
My question is: the same algorithm holds for many variables? So can i write
$$g(X_1,\ldots,X_n)=f(X_1,\ldots,X_n)\cdot q(X_1,\ldots,X_n)+r(X_1,\ldots,X_n)$$

REPLY [18 votes]: No, the polynomial division algorithm does not immediately generalize to multivariate rings. Here is a simple proof. Just as for $\rm\:\Bbb Z,\:$ a domain having an algorithm for division with smaller remainder, also enjoys Euclid's algorithm for gcds, which, in extended form, yields Bezout's identity. Therefore gcds have linear representation $\rm\:gcd(a,b) = r a + s b\:$ (i.e. Euclidean domains are Bezout). But this fails in multivariate polynomial rings $\rm\:F[x_1,\ldots, x_n],\ n \ge 2,\:$ since $\rm\:gcd(x_1,x_2) = 1\:$ but there is no Bezout equation $\rm\: 1 = x_1 f + x_2 g\ $ (evaluating  at $\rm\:x_1 = 0 = x_2\,$ $\Rightarrow$ $\rm\,1 = 0\:$ in $\rm\,F,\,$ contra field axioms).
But all Euclidean is not lost, since one can generalize the polynomial division algorithm in a way that recovers many of the important properties. For this, look up the  Grobner basis algorithm, which may be viewed as a nonlinear generalization of the Euclidean/division algorithm and Gaussian elimination.


TITLE: Stuck with this differential equation question, not getting the correct answer.
QUESTION [1 upvotes]: Suppose that, in order to buy a house, you obtain a mortgage. If the lender advertises an annual interest rate $r$, your debt $D$ will increase exponentially according to the simple O.D.E.
$$\frac{dD}{dt}=rD.$$
If you pay your debt at a rate of $P$ (continuous annual rate), the evolution of your debt will then (under assumptions of continual compounding and payment) obey the linear differential equation
$$\frac{dD}{dt}=rD−P.$$
Using this model, answer the following question:

If initial amount of the mortgage is for $400,000$, the annual interest rate is $5\%$ , and you pay at a rate of $40,000$ every year, how many years will it take you to pay off the debt?

REPLY [0 votes]: The equation is separable.
$$dt=\frac{dD}{rD-P}.$$
Then by integration,
$$t=\frac1r\log\left|\frac{rD_t-P}{rD_0-P}\right|$$ and it suffices to set $D_t=0$.


TITLE: Checking whether a map satisfies being homomorphism
QUESTION [0 upvotes]: My question might seem silly ,but excuse me for it as I've just started my hand on Ring theory.
My question is :
Like we've in case of group homomorphisms that we can check that a map $\phi :G \rightarrow H$ is in actual a homomorphism by simply checking that it satisfies the relations in the presentation of group $H$ after homomorphism.
Example:$\,\,\,\,\,$  consider group homomorphism $\phi :G \rightarrow G$ given by $\phi :g \rightarrow g^2$
If G was the group $\mathbb Z_6\times \mathbb Z_2$=$\langle a,b;a^6=1,b^2=1\rangle$  then you need to verify  that $a^{\text12}=1$ and that $b^4=1$. These  relations all clearly hold, so the map $a\rightarrow a^2,b\rightarrow b^2$ is a homomorphism.
But in case of Ring homomorphism what do we check to see whether a map satisfies being a ring homomorphism? Can anyone explain it to me with an example   .Please help....

REPLY [2 votes]: A presentation of a group $G$ can (and should be!) thought of as the choice of a free group $F$ (where the generators are our list of symbols), a group homomorphism $\pi:F\to G$, and a list of generators of $\ker{\pi}$.  In this case, we identify $G$ as $F/\ker{\pi}$.
For example, the group $G=\langle a,b;a^6=1,b^2=1\rangle$ is formed by taking $F$ to be the free group on two generators $x$ and $y$, and the generators for $\ker{\pi}$ are $x^6$ and $y^2$ (the point is that these combinations of symbols must both be sent to the identity when we replace $x$ with $a$ and $y$ with $b$).
For rings, the situation is basically identical, but the way free objects look has changed.  A presentation of a ring $R$ is a choice of a polynomial ring $F$ with coefficients in $\mathbb{Z}$.  And now, when we look at the projection $\pi: F\to R$, $\ker{\pi}$ is an ideal of $F$.
For example, the ring $\mathbb{Z}[i]$, which is the ring "generated" by $i$ with $i^2+1=0$, can be described as the quotient of $\mathbb{Z}[X]$ by the ideal $(X^2+1)$.
And your observation holds true for rings as well as groups.  If we try to define a function based on the presentations, this is like defining a map between polynomial rings.  This is a morphism of rings precisely when you can, using this map, verify the relations of the target ring using the relations of the source ring.


TITLE: Bilinear form and signature
QUESTION [0 upvotes]: I have a bilinear form such that the associate matrix is
$$A = \left(\begin{matrix} 0&0&k\\ 0&k&0\\ k&0&0\end{matrix}\right)$$
For which $k$ the quadratic for has signature $(1,2)$?

REPLY [0 votes]: This is all you need for Sylvester's Law of Inertia
$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 1 \\
 -  \frac{ 1 }{ 2 }  & 0 &  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 0 & k \\
0 & k & 0 \\
k & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 &  -  \frac{ 1 }{ 2 }  \\
1 & 0 & 0 \\
0 & 1 &  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
k & 0 & 0 \\
0 & 2 k& 0 \\
0 & 0 &  -  \frac{ k }{ 2 }  \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
0 &  \frac{ 1 }{ 2 }  &  - 1 \\
1 & 0 & 0 \\
0 &  \frac{ 1 }{ 2 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
k & 0 & 0 \\
0 & 2k & 0 \\
0 & 0 &  -  \frac{ k }{ 2 }  \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
 \frac{ 1 }{ 2 }  & 0 &  \frac{ 1 }{ 2 }  \\
 - 1 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
0 & 0 & k \\
0 & k & 0 \\
k & 0 & 0 \\
\end{array}
\right)
$$
======================================================
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right)
$$
$$  D_0 = H  $$
$$ E_j^T D_{j-1} E_j = D_j  $$
$$  P_{j-1} E_j = P_j  $$
$$ E_j^{-1} Q_{j-1}  = Q_j  $$
$$  P_j Q_j = Q_j P_j = I  $$
$$ P_j^T H P_j = D_j  $$
$$ Q_j^T D_j Q_j = H  $$
$$ H = \left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$  P_{1} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right)
$$
==============================================
$$ E_{2} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1 & 1 \\
\end{array}
\right)
$$
$$  P_{2} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 1 & 1 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
 - 1 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 1 & 0 \\
\end{array}
\right)
$$
==============================================
$$ E_{3} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 &  -  \frac{ 1 }{ 2 }  \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$  P_{3} = \left(
\begin{array}{rrr}
0 & 1 &  -  \frac{ 1 }{ 2 }  \\
1 & 0 & 0 \\
0 & 1 &  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
, \; \; \; Q_{3} = \left(
\begin{array}{rrr}
0 & 1 & 0 \\
 \frac{ 1 }{ 2 }  & 0 &  \frac{ 1 }{ 2 }  \\
 - 1 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{3} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 &  -  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
$$
==============================================
$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 1 \\
 -  \frac{ 1 }{ 2 }  & 0 &  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 &  -  \frac{ 1 }{ 2 }  \\
1 & 0 & 0 \\
0 & 1 &  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 &  -  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
0 &  \frac{ 1 }{ 2 }  &  - 1 \\
1 & 0 & 0 \\
0 &  \frac{ 1 }{ 2 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 &  -  \frac{ 1 }{ 2 }  \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
 \frac{ 1 }{ 2 }  & 0 &  \frac{ 1 }{ 2 }  \\
 - 1 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{array}
\right)
$$


TITLE: Proof involving Closure of Sets
QUESTION [1 upvotes]: Given definitions:
$ S\subseteq \Bbb R^n , S \neq \emptyset  $
The closure of $S$, denoted cl($S$), is defined as $S$  $\cup$ $S'$ where $S'$ is the set of all limit points or accumulation points of $S.$
I must show that $S'$ = cl($S'$)
Below is my proof, I'd appreciate any corrections and advice.
Proof:
In order to show $S'$ = cl($S'$), I must show double inclusion, that is $S'$ $\subseteq$ cl($S'$)  and cl($S'$) $\subseteq$ $S'$
i) $S'$ $\subseteq$  cl($S'$) because for $\forall$ $x_i \in S'$ for $i \in \Bbb N$, $x_i \in$ cl($S'$) since cl($S$) = $S \cup S'$

ii) cl($S'$) $\subseteq$ $S'$ since cl($S'$) = $S' \cup S''$ where $S''$ is the set of all limit points of $S'.$
I'm looking to see if this is makes sense mathematically. Often I have a hard time putting into words what make sense to me in my head! Also, if this proof is accurate mathematically, I would like to know if it is expressed in a clear manner.
Thanks for your insight!

REPLY [0 votes]: Since $cl(S')=S'\cup S''\supset S'$, i) is obvious.
For ii), we prove that $S''\subset S'$.
Let $x\in S''$. If $x\in S'$, then done. If not, by the definition of accumulation points, for any neighborhood $G_1$ (open set) of $x$, there is $G_1\cap (S'-\{x\})\ne \varnothing$. Pick a $y\in G_1\cap (S'-\{x\})$ and $y\ne x$.
If $y\in S$, then neighborhood of $x$ contains a point of $S$ and so $x$ is an accumulation point of $S$, or $x\in S'$, done. If not,  then there is a neighborhood $G_2$ of $y$ that  $G_2\subset G_1, \:G_2\cap (S-\{y\})\ne \varnothing$ for if $G_2$ is not a subset of $G_1$, choose $G_2=G_2\cap G_1$ instead. Pick a $z\in G_2\cap (S-\{y\})$. Then $z\in G_1$ and  $z\in S$, which means that $x$ is an accumulation point of $S$ or $x\in S'$. So $S''\subset S'$.