TITLE: holomorphic that send the unit circle to the triangle $0,1,i$
QUESTION [0 upvotes]: Let $f$ be holomorphic on the unit open disk, and continuous on the closure of that disk, such that $\forall \theta \in \mathbb{R}$ the value $f(e^{i \theta})$ lives on the boundary of the triangle with vertices $0,1,i$.
What can I say about the values that $f$ takes and lives on the triangle? For example , there is some $|z_0|<1$ , such that $f(z_0)= \frac{1}{2} (1+i) $ or  $f(z_0) = \frac{1}{10} (1+i) $? Maybe I can use Rouche, and the characterization of the elements of the convex hull, written as a linear combination. But I don't know how

REPLY [0 votes]: Non-constant holomorphic functions are continuous and open, so the boundary of the image of a domain is contained in the image of the boundary. In this case, the image of the open domain is bounded by a subset of (the boundary of) the triangle. Since the image is bounded, its boundary has to be the whole (boundary of the) triangle, so the image of the open unit disk is the open interior of the triangle.


TITLE: Equivalence relation: cardinality of quotient set
QUESTION [2 upvotes]: At $A=\mathbb{Z}^{\mathbb{N}}$ we define equivalence relation $\equiv$ by:
$$f\equiv g \iff \forall n\in \mathbb{N} ((f(2n)=g(2n)) \wedge(f(n)\cdot g(n)> 0  \ \vee f(n)=g(n)=0)) $$
a) Find the cardinality of the quotient set
b) Find the equivalence class for the constant function with value $0$
c) Decide if all equivalence classes have the same cardinality

My try:
To a) I think it's $|\mathbb{R}|$.
b) It's just $1$?
c) No since $f(n)=0$ has $1$ and if we define for $f(n)=n$, $g(2n)=2n$ and $g(2n+1)=k$ then this class is $|\mathbb{N}|$

REPLY [2 votes]: Your answers are correct, but particularly part a) could do with some elaboration. Also, your second equivalence class in c) has cardinality $|\Bbb R|$, can you see why?
The simplest way to establish a) is by estimation. On the one hand, we have $|\Bbb Z^{\Bbb N}| = |\Bbb R|$, so that $A / \equiv$ has at most this cardinality.
On the other hand, we exhibit the following set of inequivalent functions:
$$\forall g \in \Bbb Z^{\Bbb N}: f_g(n) = \begin{cases}0 &:\text{$n$ odd} \\ g(m)&:\text{$n = 2m$}\end{cases}$$
because we easily see $f_g \equiv f_h \iff g = h$. Therefore, there are also at least $|\Bbb R|$ equivalence classes, establishing the result.

Of separate interest is the determination of all equivalence classes. It is not too hard to see that on odd $n$, two functions $f$ and $g$ being equivalent amounts to:
$$\operatorname{sgn}(f(2n+1)) = \operatorname{sgn}(g(2n+1))$$
for all $n$, where $\operatorname{sgn}(n)$ is $0, 1$ or $-1$ according to whether $n$ is zero, positive or negative. It follows that we can describe all equivalence classes by the functions $f_{g,s}$ (where $g \in \Bbb Z^{\Bbb N}$ and $s \in \{0,1,-1\}^{\Bbb N}$) defined as:
$$f_{g,s}(n) = \begin{cases}g(m) &: n = 2m\\s(m) &: n = 2m+1\end{cases}$$


TITLE: Understanding counting.
QUESTION [3 upvotes]: I encountered this question recently:

Suppose there are 3 benches in the front row and 7 benches in the
  second row, how many ways a group of 10 children can be seated in such
  arrangement?

There were given two solutions to the same problem:

Its an Arrangement hence there are $10!$ ways. The solution seem easy to understand.
The other solution was like:

We can choose $3$ children for the first row in $10\choose 3$ ways, those $3$ can be arranged in $3!$ ways and the remaining $7$ can be arranged in $7!$ ways in the back row. So, the total number of ways are ${10 \choose 3}*3!*7!$ which equals $10!$.
I seriously got bowled over by the second solution, it went over my head like anything.
I am new to discrete math and as far as I know counting boils down to two simple rules sum and product. Both of the above solutions are using the same product rule but why I am failing to understand the second one?
Question
I know the product rule but still the multiplication of the sub-parts is not making sense to me. How, to understand or get better mental model of it?

REPLY [2 votes]: The rule of multiplicatinn is simple.   If there are $x$ ways to perform a subtask and $y$ ways to perform a second (independent) subtask, and the tasks are performed in serial then there are $xy$ ways to perform whole task.
Think of it as branching paths.   If the path branches into $x$ paths and then each of these branches into $y$, how many ways are they to take?
Similarly if you take a bushwalk which has $x$ paths to get from $A$ to $B$ and there are $y$ paths from $B$ to $C$, then how many different ways are there to go from $A$ to $B$ to $C$?
When independent subtasks are performed in serial then multipling the counts of subtasks obtains the count of the whole task.
[Serial: both tasks must be performed; Parallel: the tasks are alternatives.]

We write $3!$ as the count of ways to arrange three distinct items; and likewise $7!$ and $10!$ are the ways to arrange seven, or ten, distinct items, respectively.
Well, now we write $\binom {10}3~$, or sometimes $~^{10}\mathbb C_3$, to represent the count of ways to choose three from ten distinct things.
But what exactly is this $\binom {10}{3}$ notation?   How do we find its value?
You have examined two ways to count how to arrange ten children among ten seats; when the seats are in two rows; three and seven.
Because these are ways to perform the exact same task; the counts must be the same.

You have counted the ways to select three of ten children, then arrange them among three seats, and finally arrange the remaining children among seven seats.   This is: $~\binom{10}3\cdot 3!\cdot 7!$.
You have also counted the ways to simply arrange all ten children among all ten seats.   That is: $~10!~$.

Hence we find that: $\dbinom{10}3 = \dfrac{10!}{3!\,7!}$ .

In general we find that $\dbinom{n}{r} ~=~ \dfrac{n!}{r!~(n-r)!}$ for all non-negative integer $r,n$ such that $0\leq r\leq n$
You use verify this with smaller examples: What are $\binom 3 1$ and $\binom 4 2$ ?


TITLE: Example of an elastic transverse wave
QUESTION [0 upvotes]: I read in my student's book that elastic waves can be transverse and longitudinal. After a little research I knew that there is an elastic solid (not an elastic wave).
I can imagine a longitudinal wave (the horizontal spring example), but was interested in the case of transverse elastic wave.
Here, "Transverse waves commonly occur in elastic solids", I want to see the effect of elasticity in a transverse wave. The easiest example to me of a transverse wave is the experiment of a rope, fixed in one end, and in the other end moved up and down fast vertically (back to the original horizontal level). But this is not applicable for the concept of elasticity, since each set of particles of the rope are due to constantly (continuously) driving force which finally sets the particles back to their original calm horizontal level with a null force (as far as I can imagine), so the question of elasticity doesn't hold here since it deals with the case of what comes after a non-null force with deformation of the object.
Can someone give an example? Is the book's sentence correct ?

REPLY [0 votes]: As @anna v pointed out in the comment, according to the definition in this link,

the displaced particles transfer momentum to adjoining particles, and are themselves restored to their original position.

I was considering this force that restores them back to their original position as a foreign force, this is why I was wondering where the elasticity is. The source of confusion was that this momentum transfer incident was itself the cause of wave propagation (not only was it the restoring cause), which was a little non-evident with the definition of elasticity, where the cause or force is usually an outsider one.


TITLE: Universal coefficient theorem for local ring
QUESTION [1 upvotes]: Let $R$ be a commutative local artin $k$-algebra,where $k$ is a field with characteristic $0$.I wonder whether universal coefficient theorem holds in this case.Namely,if $C$ is a chain of flat $R$-modules. Then does $H(C\otimes_R S)\cong H(C)\otimes_R S$ whenever $H(C)$ is flat $R$-module. $H(C)$ is homology of chain complex $C$.$S$ is commutative ring over $R$
I checked wikipedia and nLab,it seems that there is only the case when $R$ is field or Principal ideal domain.I dont know whether one can use so called Kunneth spectral sequence to make arguments.I just started to understand
Thanks

REPLY [4 votes]: It doesn't hold. Take $k$ any field (of characteristic zero if you wish), $R=k[\epsilon]/(\epsilon^2)$ its ring of dual numbers, $S=R/(\epsilon)=k$ its residue field, and $C$ the periodic complex
$$\cdots\rightarrow R\stackrel{\epsilon}\longrightarrow R\rightarrow\cdots.$$
This is a complex of projective (hence flat) $R$-modules with trivial (hence flat) homology $H(C)=0$, so $H(C)\otimes_RS=0$, but $C\otimes_RS$ is
$$\cdots\rightarrow S\stackrel{0}\longrightarrow S\rightarrow\cdots,$$
hence $H_n(C\otimes_RS)=S\neq 0$ in each degree $n\in\mathbb Z$.


TITLE: Show that a subspace $U$ exists such that the orthogonal projection of $v$ onto it has the specified length $\alpha < \|v\|$
QUESTION [1 upvotes]: Given the inner product space $V$ such that $\dim V > 1$.
Show that for every $\alpha \in \mathbb R$ and $v \in V$ such that $0\le \alpha \le\| v\|$ there exists a subspace $U\subset V$ such that the orthogonal projection of $v$ onto $U$ has length $\alpha$.

I think this has something to do with representing $U$ in an orthonormal basis $\{u_1\}$ for example and then saying $P_U(v)=\langle v,u_1\rangle v$. But I'm not really sure about this.

REPLY [3 votes]: As in my comment, let $w$ be a nonzero vector which is perpendicular to $w$. Such a $w$ exists; take any nonzero $x$ which is not a scalar multiple of $v$ (which exists since $\dim V>1$), and let $w=x-\frac{\langle{x,v}\rangle}{\langle{v,v}\rangle}v$.
Recall that for any $u\neq 0$, the length of the projection of $v$ onto the subspace generated by $u$ is
$$
\frac{|\langle u,v\rangle |}{\|u\|}
$$
Now, let $u=tw+(1-t)v$. Using the above formula, the length of the projection is
$$
\frac{|\langle tw+(1-t)v,v\rangle |}{\|tw+(1-t)v\|}=\frac{|\langle tw+(1-t)v,v\rangle |}{\sqrt{\langle tw+(1-t)v,tw+(1-t)v\rangle}}=\frac{|1-t|\cdot \|v\|^2}{\sqrt{t^2\|w\|+(1-t)^2\|v\|^2}}
$$
Now, consider the above as a function of $t$. We notice three things:

When $t=0$, the function is $\|v\|$.
When $t=1$, the function is $0$.
The function is continuous in $t$.

By the intermediate value theorem, for every $0<\alpha<\|v\|$, there exists a $t$ for which the value of the function (the length of the projection) is equal to $\alpha$, as desired.


TITLE: $\frac {\partial}{\partial t}T$ vs $\frac d{dt} T$.
QUESTION [4 upvotes]: Suppose we have a function $T_1=F(x,y,t)$. Now suppose that $x=g(t),y=h(t)$, so we have a new $T_2=F(x(t),y(t),t)$, so then we have that $\frac \partial{\partial t} T_2=F_t$ and $\frac d{dt}T_2=F_x x_t+F_yy_t+F_t$.
My question is, if we let $x,y$ and $t$ be independant variables, does the symbol $\frac d{dt}T_1$ (if we use the same idea as before, we should have that $\frac d{dt}T_1=\frac {\partial}{\partial t}T_1$, is this true?) have any meaning?

REPLY [1 votes]: $F(x,y,t)$ is a function of three variables. So you can compute $F_x,F_y$ and $F_t$ (assuming they exist).
If $x=x(t)$ and $y=y(t)$ then $F(x(t),y(t),t)$ is a function of one variable: $t.$ You can compute its derivative $\frac{d}{dt}F$ using that chain rule. As you have said $$\dfrac{d}{dt}T=F_xx'+F_yy'+F_t.$$ So, if $x$ and $y$ are constant (don't depend on $t$) then
$$\dfrac{d}{dt}F=F_xx'+F_yy'+F_t=0F_x+0F_y+F_t=F_t=\frac{\partial}{\partial t}F.$$
Going back to your question, $\dfrac{d}{dt}F$ is the derivative of a function of one variable and thus, if $x,y$ and $t$ are independent then $\dfrac{d}{dt}F$ makes no sense. $\dfrac{d}{dt}F$ can be thought as the derivative along a curve (or a directional derivative in the direction of the tangent of the curve). So, if $x,y$ and $t$ are independent then we don't have any given direction to get the derivative.


TITLE: Function Composition Formulas
QUESTION [0 upvotes]: The a's and b's are confusing me!
For a), I have deduced that g(x) would be something like $-x+a$ where $a$ = the output of $f(x)$... but then you couldn't write g(x) in terms of x and a... ...so how do I write $h(x)$ from this information?
so f(x) has the point (b,a) and g(x) has the points (0,a) and (a,0)... confusing to me...

REPLY [1 votes]: $g$ is a straight line with gradient $-1$, and $y$ intercept $a$, so its equation is $g(x)=a-x$. $f$ looks like a quadratic, with minimum at $(0,0)$ and a point at $(b,a)$, so $f(x)=\frac{ax^2}{b^2}$. Do you think you can compute the compositions from this?


TITLE: Proving $A,B \in M_n \mathbb{C}$ are similar and finding an $X$, such that $A = X^{-1} B X$
QUESTION [1 upvotes]: If I have two matrices $ A,B \in M_n \mathbb{C}$, what are the necessary and sufficient conditions for $A$ to be similar to $B$.
I know that they must share properties such as rank, characteristic polynomial, Frobenius normal form etc. Does the full list provided here contain all the sufficient conditions for $A$ and $B$ to be similar?
Furthermore, if you know that $A \sim B$, how does one practically calculate an $X \in GL_n \mathbb{C}$, satisfying
$$A = X^{-1} B X$$

REPLY [1 votes]: Necessary and sufficient is that they have the same Jordan normal form.
In practice, you can look at $XA - B X = 0$ as a system of linear equations in the entries of $X$, and solve it.  A "generic" solution will be invertible.


TITLE: Can we realize a distance minimizer via converging sequences?
QUESTION [2 upvotes]: Let $X$ be a Riemannian manifold*, and $S$ a compact submanifold of $X$.
Assume there exists an open, dense subset $Y$ of $\,X$, such that for any element $y \in Y$, there exists a unique element in $S$ closest to $y$; i.e there is a function $\tilde s:Y \to S$ such that $$ d(y,\tilde s(y))=d_S(y)=\inf\{d(s,y)|\, \,s \in S\},$$
and $\tilde s(y)$ is the only element in $S$ satisfying the above equality.
Question:
Let $x \in X$, and let $\tilde s \in S$ be a closest element to $x$ in $S$. (such an element exists by compactness of $S$).
Is it true that there exists a sequence $y_n \in Y$, $y_n \to x$, such that $\tilde s(y_n) \to \tilde s$?
Note that I do not assume $x$ has a unique closest point on $S$. (If this were true the question would become trivial; On any subset consisting of elements which have unique closest points, the "closest point mapping" $\tilde s$ is continuous).

*(Actually, one could ask the question in a more general setting when $X$
is a metric space, and $S$ is a subset of it, but I am not sure which "strange pathologies" can arise then)

REPLY [0 votes]: I am adding some details to the answer of HK Lee:
We assume $X$ is geodesically convex, i.e there is a minimizing geodesic between every two points. (In particular, our argument holds for any complete manifold).
Let $x \in X$, and suppose $s_0$ is a closest point to $x$ in $S$. Let $c:[0,1] \to X$ be a minimizing geodesic from $x$ to $S$, i.e $c(0)=x,c(1)=s_0$.
We begin with the following observation:
Lemma (1): For every $0<t_0<1$, $s_0$ is the a unique closest point to $c(t_0)$ on $S$.
Proof Lemma (1):
The fact $s_0$ is a closest point to $c(t_0)$ is trivial.
Assume $s_1 \neq s_0$ satisfies  $$d(c(t_0),s_1) = d(c(t_0),S) = d(c(t_0),s_0),\ s_1\in S$$
Let $\gamma$ be a minimizing geodesic from $c(t_0)$ to $s_1$.
Look at the concatentation $\gamma * c|_{[0,t_0]}$; $$L(\gamma)=d(c(t_0),s_1)= d(c(t_0),s_0)=L(c|_{[t_0,1]}),$$ so
$$L(\gamma * c|_{[0,t_0]})=L(c|_{[0,t_0]}) + L(\gamma)=L(c)=d(x,s_0)=d(x,S)$$
$\gamma * c|_{[0,t_0]}$ is therefore a broken path (not smooth at $t_0$) connecting  $x$ to $s_1$. Since a minimizing path between every two points must be smooth, it follows that $ \gamma * c|_{[0,t_0]}$ is not a minimizer, i.e $d(x,s_1) < L(\gamma * c|_{[0,t_0]})=d(x,S)$ which is a contradiction.

Lemma (2): For every $0<t<1$, there exists a sequence $y_n \in Y$ such that $y_n \to c(t), \tilde s(y_n) \to s_0$.
Proof Lemma (2):
Let $0<t<1$, and let $y_n \in Y, y_n \to c(t)$. $$(1) \lim_{n \to \infty} d(y_n,\tilde s(y_n))=\lim_{n \to \infty} d(y_n,S)=d(c(t),S)$$
Since $S$ is compact, we can assume W.L.O.G that $\tilde s(y_n)$ is converging to some $\tilde s \in S$. Thus,
$$ (2) \lim_{n \to \infty} d(y_n,\tilde s(y_n))=d(c(t),\tilde s) $$
$(1),(2)$ imlpies $d(c(t),S)=d(c(t),\tilde s)$
Lemma (1) implies $\tilde s=s_0$.

Proof of the main proposition:
Define $t_n=\frac{1}{n}$. By Lemma (2) (applied for each $t_n$), there exists $y_n \in Y$ such that $d(y_n,c(t_n))<\frac{1}{n}$, and $d(\tilde s(y_n),s_0) < \frac{1}{n}$. Thus, $y_n \to x, \tilde s(y_n) \to s_0$, as required.


TITLE: Does $f(x)\leq g(x)$ imply $\lim_{x\to 0_+} f(x)/x \leq \lim_{x\to 0_+} g(x)/x$
QUESTION [1 upvotes]: Let $f,g:[0,\infty)\mapsto[0,\infty)$ be convex functions such that $f(x)\leq g(x)$ for all $x\in [0,\infty).$ Suppose that both
$ \lim_{x\to 0_+} \frac{f(x)}{x}$
and
$ \lim_{x\to 0_+} \frac{g(x)}{x}$ are well defined and finite. Is it true that
$$ \lim_{x\to 0_+} \frac{f(x)}{x} \leq \lim_{x\to 0_+} \frac{g(x)}{x}?$$
It follows immediately that $ \lim_{x\to 0_+} {f(x)} \leq  \lim_{x\to 0_+} {g(x)}$ but the factor of $1/x$ in the above throws me off. Any proofs/guidance/counter examples would be appreciated.

REPLY [1 votes]: I am a student too so i hope this is correct.
Let note:$\lim_{x\to 0_+}f(x)/x=l_f\geq 0$ and $\lim_{x\to 0_+}g(x)/x=l_g \geq 0$ Because by assumption $f(x)\geq$ and $g(x)\geq0$ for $x \geq 0$
By def we can writte: $\forall \epsilon>0 , \exists\delta_f>0 \; s.t. \; \forall x\in(0; \delta_f) \Rightarrow |\frac{f(x)}{x}-l_f|<\epsilon \;or\; -\epsilon+l_f<\frac{f(x)}{x}<\epsilon+l_f$
Same for $g(x)/x$
Now let suppose that $l_f>l_g$ so by def $\exists \epsilon_0=\frac{l_f-lg}{4}>0$ s.t:
$\exists\delta_{f0}>0 \; \forall x\in(0; \delta_{f0}) \Rightarrow |\frac{f(x)}{x}-l_{f0}|<\epsilon_0$
$\exists\delta_{g0}>0 \; \forall x\in(0; \delta_{g0}) \Rightarrow |\frac{g(x)}{x}-l_{g0}|<\epsilon_0$
So we get for: $\delta_{min}=min(\delta_{f0},\delta_{g0}) \Rightarrow \forall x\in(0; \delta_{min})$  we have $|\frac{f(x)}{x}-l_{f}|<\epsilon_0$ and $|\frac{g(x)}{x}-l_{g}|<\epsilon_0$
So: $\frac{g(x)}{x}<\epsilon_0+lg=\frac{l_f+3l_g}{4}$ and $\frac{f(x)}{x}>l_f-\epsilon_0=\frac{3l_f+l_g}{4}$.
CONTRADICTION! because in this case we get: $\forall x\in(0; \delta_{min})$ the following inequality: $ \frac{g(x)}{x}<\frac{f(x)}{x} \Rightarrow g(x)<f(x)$ (as x>0).
Because: $\frac{l_f+3l_g}{4}<\frac{3l_f+l_g}{4}\Leftrightarrow l_f+3l_g<3l_f+l_g \Leftrightarrow l_f-l_g>0$. By absurd assumption we supposed $l_f>l_g$
Q.E.D


TITLE: Snell's law for classical particle?
QUESTION [0 upvotes]: I was thinking whether there is any analogous Snell's Law for classical particles.
for example I have a problem where a particle (in 2D) crosses a boundary separating potential $U_1$ and $U_2$. It enters the boundary with angle $\theta_1$ with the normal to the boundary.
I wanted to find the $\theta_2$ that the particle makes with the boundary when it has entered the new region.
If there isnt any Snell's law for classical particles, How to solve this problem ??
Are the energy and momentum conservation equations enough?

REPLY [1 votes]: Newton has done this for you!
He used his ideas about Mechanics and assumed that the component of velocity of the corpuscles (the name he gave to the particles of light) perpendicular to the interface, as they moved from one medium into another medium, changes due to the change in the potential, whereas the components of velocity parallel to the interface did not change as there were no forces in that direction, so $v_1\sin \theta_1= v_2\sin \theta_2$.
So the refractive index $=\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_1}{v_2}$ which from your diagram means that the speed of light is less in medium $2$ than in medium$1$ which  was later shown experimentally to be incorrect.


text:
\section{Overview}

Solving a polynomial equation calls for a means to overcome the polynomial's symmetry.  In the case of the fifth-degree equation, the general symmetry group is the symmetric group \sym{5}.  In terms of Galois theory, we can reduce the symmetry to that of the alternating group \alt{5} by adjoining the square root of the polynomial's discriminant to the coefficient field.  Our reward for this reduction is that we can realize \alt{5} as the rotational symmetries of the regular icosahedral configuration of the complex projective line \CP{1}---that is, the Riemann sphere.

By exploiting icosahedral structure, Doyle and McMullen constructed a quintic-solving algorithm at the core of which is a map $\phi$ that respects the \alt{5} symmetry.\cite{dm}  The map is strongly critically finite, meaning that it's critical set \CC{\phi}, the twenty face-centers of the icosahedron, is $\phi$-invariant; that is, $\phi(\CC{\phi})=\CC{\phi}$.  In particular, each superattracting critical point has period two.  It follows that almost every point in \CP{1} belongs to the basin of attraction of some two-cycle in \CC{\phi}.  The procedure employs $\phi$'s dynamics in a way that partially breaks the \alt{5} symmetry and, with one iterative run, computes two roots.

Recent work determined all icosahedrally-symmetric maps with \emph{internally periodic} critical sets of size $60$.\cite{soccer,icos31}  Internal periodicity means that the map acts on its critical set as a permutation.  Here, we build a quintic-solving device around the dynamics of one such map $g$ whose critical points have period five.  Since the superattracting set has generic size, the dynamics of $g$ effectively breaks \emph{all} of an equation's \alt{5} symmetry.  Accordingly, the algorithm produces all five roots with a single iterative run.

Computational results are produced by \emph{Mathematica} and basins-of-attraction plots are the product of \emph{Dynamics 2}. \cite{dyn2}

config: arxiv


TITLE: Proving a Sequence Does Not Converge
QUESTION [1 upvotes]: I have a sequence as such:

$$\left( \frac{1+(-1)^k}{2}\right)_{k \in \mathbb{N}}$$

Obviously it doesn't converge, because it alternates between $0,1$ for all $k$. But how do I prove this fact?
More generally, how do I prove that a sequence does not converge? Are there any neat ways other than "suppose, for contradiction, that the sequence converges. Then..."?

REPLY [1 votes]: $\{a_k\}$ converges if and only if $\{a_{2k}\}$ and $\{a_{2k+1}\}$ converge to the same limit. Here
$$
a_{2k}=1\rightarrow1, \quad a_{2k+1}=0\rightarrow 0.
$$


TITLE: CAR on fermionic Fock spaces
QUESTION [1 upvotes]: I'm trying to prove by myself the canonical commutation relations (CAR):
$$[a(\varphi),a(\psi)] = 0 \quad \mbox{and} \quad [a^{\dagger}(\varphi),a^{\dagger}(\psi)] = 0$$
on fermionic Fock spaces. Here $[A,B] := AB+BA$ and $\varphi, \psi$ are elements of a Hilbert space $\mathcal{H}$. The Fock space is given by $\mathcal{F}(\mathcal{H}) := \bigoplus_{n=0}^{\infty}A_{n}\mathcal{H}$ where $A_{n}$ is the antisymmetrization operator. I'm trying to address the first commutation relation.
Now, I know that:
$$a(\varphi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) = \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \varphi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n})$$
where $\hat{\varphi}_{j}$ means omit this entry.
Thus, my reasoning is the following:
$$ a(\varphi)a(\psi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) = a(\varphi) \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n})$$
Now, using the linearity of $a(\varphi)$ we get:
\begin{align}
a(\varphi)a(\psi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) &= a(\varphi) \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n}) \\ &=  \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle a(\varphi) A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n}) \\ &= \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle \frac{1}{\sqrt{n-1}}\sum_{k=1}^{n-1}(-1)^{k+1}\langle \varphi, \varphi_{k}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \hat{\varphi}_{k}\otimes \cdots \otimes \varphi_{n})
\end{align}
And we get the same expression but exchanging $\psi$ and $\varphi$ if we evaluate $a(\psi)a(\varphi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n})$. Now, the sum of these terms is supposed to be zero, which is not obvious to me but it probably is.
Is my reasoning correct? Is there an easier way to approach this?

REPLY [1 votes]: There is a slight but significant error in your last term. The sum over $k$ should go from $1$ to $n$ excluding $j$, not from $1$ to $n-1$, but more importantly the sign should be $(-1)^{k+1}$ if $k<j$ and $(-1)^{k}$ if $k>j$. So:
$$a(\varphi)a(\psi) [A_n(\varphi_1\otimes .... \varphi_n)] = \frac1{\sqrt n}\frac1{\sqrt{n-1}}\left[ \begin{split} \sum_{j=2\\ k<j}^n (-1)^{j+k} \langle \psi,\varphi_j\rangle\langle\varphi,\varphi_k\rangle A_{n-2}(j,k)\\+ \sum_{j=1\\ k>j}^{n-1} (-1)^{j+k+1} \langle \psi,\varphi_j\rangle\langle\varphi,\varphi_k\rangle  A_{n-2}(j,k) \end{split}\right]$$
If you look at $a(\psi)a(\varphi) A_n(\varphi_1\otimes...\otimes \varphi_n)$ the only thing that changes is that $\varphi$ is contracted with $\varphi_j$ and $\psi$ with $\varphi_k$ in the inner products. This can be corrected by substituting $j$ with $k$ and vise versa. If you do this the sum with $k<j$ swaps with the sum with $k>j$. But now the signs don't match, there is an extra factor $-1$. So $a(\varphi)a(\psi)=-a(\psi)a(\varphi)$.


text:
\begin{document}

\begin{abstract}
The aim of this note is to investigate the relation between two types of
non-singular projective varieties of Picard rank 2, namely the Projective
bundles over Projective spaces and certain Blow-up of Projective  spaces.
\end{abstract}

\maketitle
{\bf Keywords:} Projective spaces; Projective bundles; Blow-ups.

\section{Introduction} \setcounter{page}{1}

Let $X$ be the projective variety obtained from projective space
$\PP^n= \PP((\CC^{n+1})^*)$ by blowing up along a
linear subspace $\PP(W^*)$ of co-dimension $r,$  where $\CC^{n+1} \to W$ is
a quotient vector space of dimension $n-r+1.$ If $K$ is the kernel of
surjective linear map $\CC^{n+1} \to W$
resulting projective variety admits a projective bundle
structure over the smaller dimensional projective space $\PP(K^*).$
In fact
$$ X \simeq \PP(W\otimes\OOO_{\PP(K^*)}\oplus\OOO_{\PP(K^*)}(1)).$$

Question: Are there any other examples of
blow-ups of a projective space along a sub variety
that also have a structure of
a projective bundle over a projective space?

Nabanita Ray in \cite{NR} gave examples of non-linear sub varieties in
$\PP^i, i=3,4,5$  whose blow-ups are
projective bundles over  $\PP^2.$

In this note we provide several other examples of  Projective bundles over a Projective space $\PP^n$
which can be realised as a blow up of some $\PP^m$ along a non-linear sub variety.

\begin{thm}\label{thm1}
	Let $n\geq 2$ be an integer and $N= n^2+n-1.$  Let
	$$E = \underline{Hom}(\OOO_{\PP^n}^n, T_{\PP^n}(-1))$$
	be the rank $n^2$ vector bundle on $\PP^n,$ where $T_{\PP^n}(-1)$ is tangent bundle of $\PP^n$ twisted by the inverse of the ample bundle $\OOO_{\PP^n}(1).$
	The projective bundle $\PP(E)$ is isomorphic to a blow-up of
	$\PP^N=\PP(Hom(\CC^n,\CC^{n+1}))$
	along the sub variety of all linear mappings of rank at most $n-1.$
\end{thm}

\begin{thm}\label{thm2}
	Let $n$ be an even integer and $N= \frac{n(n+1)}{2}-1.$ Let $V= \wedge^2(T_{\PP^n}(-1))$ be the rank $N+1-n= \frac{n(n-1)}{2}$
	vector bundle over $\PP^n.$ The projective bundle $\PP(V)$ is isomorphic to
	$\PP^N=\PP(Alt({n+1}))$ blown-up
	along the sub variety  consists of all alternating linear mappings of rank at most
	$n-1,$ where $Alt({n+1})$ denote the space of all alternating linear mappings from
	$\CC^{n+1}$ to itself.
\end{thm}

\section{The Projective bundle  $\PP(\underline{Hom}(\OOO_{\PP^n}^n, T_{\PP^n}(-1)))$}

{\bf Proof of Theorem(\ref{thm1}):} On $\PP^n$ by applying
$$\underline{Hom}(\OOO_{\PP^n}^n,-) $$
to the standard exact sequence
\begin{equation}\label{eq01}
 0 \to  \OOO_{\PP^n}(-1) \to
\OOO_{\PP^n}^{n+1} \to T_{\PP^n}(-1) \to 0.
\end{equation}
we obtain the exact sequence
\begin{equation}\label{eq0}
 0 \to \underline{Hom}(\OOO_{\PP^n}^n, \OOO_{\PP^n}(-1)) \to
\underline{Hom}(\OOO_{\PP^n}^n, \OOO_{\PP^n}^{n+1}) \to E \to 0.
\end{equation}

Note that $E$ is a globally generated vector bundle and ${\rm dim}H^0(E)=n(n+1).$
Hence we obtain a morphism from
$$\phi : \PP(E) \to \PP^{n(n+1)-1}.$$
 We claim that this is a surjective morphism of degree one. i.e., $\phi$ is a
 bi-rational morphism and hence $\PP(E)$ is a blow-up of $\PP^{n(n+1)-1}$
 [See, Chapter II, Theorem 7.17. cite{Ha}]
  Let $\xi=\OOO_{\PP(E)}(1)$ denote the tautological line bundle on  $\PP(E).$
  Then there is an exact
  sequence of vector bundles on $\PP(E)$
\begin{equation}\label{eq1}
 0 \to \Omega^1_{\PP(E)/{\PP^n}} \to \pi^*(E) \to \xi \to 0,
\end{equation}
where $\pi : \PP(E) \to \PP^n$ is the natural projection.
Since $\xi^{(n(n+1)-1)}$ is the degree of the morphism $\phi$ it is enough to
show that $\xi^{(n(n+1)-1)}=1.$
Since $E$ is vector bundle of rank $n^2$ in the cohomology ring of
$\PP(E)$ the relation
\begin{equation}\label{eq2}
 \xi^{n^2} = \sum_{i=1}^{n}(-1)^{i+1}\pi^*(C_i(E))\xi^{n^2-i}
\end{equation}
holds, where $C_i(E),\, (1\leq i \leq n)$ are the Chern classes of $E.$
Repeated use of the relation \ref{eq2}  gives
\begin{equation}\label{eq3}
\xi^{n^2+n-1} = (-1)^n\pi^*(S_n(E))\xi^{n^2-1},
\end{equation}
where $S_n(E) $ is the $n$th Segre class of $E.$
By equation (\ref{eq1}) and the fact that the  total Segre class is the
inverse of the total Chern class we deduce that
$S_n(E) = (-1)^n h^n,$
where $h$ is the class of the line bundle $\OOO_{\PP^n}(1)$ in ${\rm H}^2(\PP^n, \ZZ).$
Hence $\xi^{n^2+n-1} = 1.\pi^*(h^n)\xi^{n^2-1}.$ Since $\pi^*(h^n)\xi^{n^2-1}$ is
the class of a point in $\PP(E)$ it follows that the degree of the map $\phi$ is one.
$\hfill{\Box}$

{\bf Remark:} {\it The bi-rational morphism $\phi$  can be described geometrically
as follows:
   let
   $$ T = \{ (v, \varphi)\in \PP^n \times \PP({\rm Hom}(\CC^{n},\CC^{n+1}))| (\varphi)^t(v)=0 \},$$
   where $(\varphi)^t :\CC^{n+1} \to \CC^n$ is the transpose of the map
   $\varphi :\CC^{n} \to \CC^{n+1}.$
   Then $T \simeq \PP(E)$  and the bi-rational morphism $\phi$ is the projection
   onto the second factor. Let
   $$D= \{\varphi \in \PP({\rm Hom}(\CC^{n},\CC^{n+1}))|\,\, \text{rank}(\varphi)\leq n-1\},$$
   $F = \phi^{-1}(D)$ and $U= \PP(E)\setminus F.$
The morphism $\phi$ is one-one on the open set $U.$ }

Next we prove that the set $F$ in the previous Remark is a divisor and we identify
its class in the Picard group of $\PP(E).$ Note that Picard group of $\PP(E)$
is equal to ${\rm H}^2(\PP(E),\ZZ) = \ZZ[\pi^*(h)] \oplus \ZZ [\xi].$

\begin{lem}\label{lem1}
	With the notations of previous Remark the set $F$ is a divisor in $\PP(E)$
	and its divisor class is $ n[\xi]-[\pi^*(h)].$
\end{lem}

{\bf Proof:}  Rewriting the exact sequence (\ref{eq0}) as
\begin{equation}\label{eq4}
0 \to \OOO_{\PP^n}(-1)^n \to \OOO_{\PP^n}^{n(n+1)} \to E \to 0
\end{equation}
and taking $n$-th symmetric power
we obtain a long exact sequence
\begin{equation}\label{eq5}
0 \to \OOO_{\PP^n}(-n)\to \OOO_{\PP^n}^{n(n+1)} \otimes \wedge^{n-1}(\OOO_{\PP^n}(-1)^n)\to \cdots
\end{equation}
\begin{equation}\label{eq6}
\cdots \to S^{n-1}(\OOO_{\PP^n}^{n(n+1)})\otimes \OOO_{\PP^n}(-1)^n \to S^{n}(\OOO_{\PP^n}^{n(n+1)}) \to S^n(E )\to 0.
\end{equation}
Tensoring the long exact sequence (\ref{eq5}) (\ref{eq6}) by $\OOO_{\PP^n}(-1)$
 and computing the cohomology
we deduce
$${\rm H}^0(\PP(E), \xi^{\otimes n}\otimes \pi^*(\OOO_{\PP^n}(-1)))\simeq {\rm H}^0(S^n(E)(-1)) $$
$$ \simeq {\rm H}^n(\OOO_{\PP^n}(-n-1)) \simeq \CC.$$
This proves that the divisor class $ n[\xi]-[\pi^*(h)]$ contains a unique effective divisor.
From the exact sequence (\ref{eq0}) we deduce that the Segre class $S_{n-1}(E)$
is equal to $(-1)^{n-1}nh^{n-1}$ and hence
$(n[\xi]-[\pi^*(h)]).\xi^{n^2+n-2}= n\pi^*(h^n)\xi^{n^2-1}-(-1)^{n-1}\pi^*(h)\pi^*(S_{n-1}(E))\xi^{n^2-1}= 0.$
For $v \in \PP^n$
the fibre  $E_v$ is isomorphic to ${\rm Hom}(\CC^n, {\CC^{n+1}/{\CC.v}}).$ For
$\varphi \in {\rm Hom}(\CC^n, \CC^{n+1})$  rank of $(\varphi)^t$ is less than or equal to $n-1$
if and only if image of $\bar{\varphi}\in {\rm Hom}(\CC^n, {\CC^{n+1}/{\CC.v}})$ of $\varphi$
has rank less than or equal to $n-1,$ i.e., $\bar\varphi$ is not an isomorphism.
 Since the complement of
isomorphisms in ${\rm Hom}(\CC^n, {\CC^{n+1}/{\CC.v}})$ is given by vanishing of the
homogeneous polynomial of degree in $n,$ i.e., section of $\OOO_{\PP(E_{v})}(n).$
The set
$$ F_v =\{\bar{\varphi} \in {\rm Hom}(\CC^n, {\CC^{n+1}/{\CC.v}})|  {\rm rank}(\bar\varphi) \leq n-1 \} $$
is an irreducible divisor given by the vanishing of the restriction to  $\PP(E_v)$ of the
non-zero section (unique upto non zero scalar) of
$${\rm H}^0(\PP(E), \xi^{\otimes n}\otimes \pi^*(\OOO_{\PP^n}(-1)))$$
The non-zero section $t$ (unique upto multiplication by a
non zero scalar) determines a section of $\OOO_{\PP(E_{v})}(n).$ It is clear that $F_v$ is the
set of zeros of  the section $t|_{\PP(E_v)}.$  This proves that the set $F$ is the support of the
divisor the $n[\xi]-[\pi^*(h)].$             $\hfill{\Box}$

{\bf Remark 1:} {\it Theorem (\ref{thm1}) can be used to obtain  for
blow up of $\PP^m, \,(2n-1 \leq m \leq n(n+1)-1),$  along a non linear sub-variety a projective
bundle structure over $\PP^n$ for any integer $n \geq 2.$}

\section{Projective bundle $\PP(\wedge^2(T_{\PP^n}(-1))).$}

{\bf Proof of Theorem(\ref{thm2}):}

  Let  $n \geq 3$ be an integer.  By taking 2nd exterior  power in the exact sequence (\ref{eq01})
  yields the sequence
  \begin{equation}\label{eq7}
 0 \to  T_{\PP^n}(-2) \to
\wedge^2\OOO_{\PP^n}^{n+1} \to \wedge^2(T_{\PP^n}(-1)) \to 0.
\end{equation}
If we identify ${\rm H}^0(\OOO_{\PP^n}^{n+1})$ with $\CC^{n+1}$ then the
${\rm H}^0(\wedge^2\OOO_{\PP^n}^{n+1})$ gets identified with $Alt({n+1}),$
where $Alt({n+1})$ denote the set of all alternating linear mappings from
	$\CC^{n+1}$ to itself.  Using the exact sequence (\ref{eq7}) we obtain
	$$H^0(\wedge^2(T_{\PP^n}(-1))) \simeq Alt({n+1}).$$
	Since $T_{\PP^n}(-1)$ is globally generated $V= \wedge^2(T_{\PP^n}(-1))$  is generated by
$Alt({n+1}).$
 Therefore we get a morphism
 \begin{equation}\label{eq8}
 \psi : \PP(V) \to \PP(Alt({n+1})).
 \end{equation}
 We claim that $\psi$ is bi-rational i.e., degree of $\psi$ is one and hence
  $\PP(V)$ is a blow-up of $\PP(Alt({n+1})).$
 [See, Chapter II, Theorem 7.17. cite{Ha}]
  Let $\zeta=\OOO_{\PP(V)}(1)$ denote the tautological line bundle on  $\PP(V).$
  Then there is an exact
  sequence of vector bundles on $\PP(V)$
\begin{equation}\label{eq9}
 0 \to \Omega^1_{\PP(V)/{\PP^n}} \to \pi^*(V) \to \zeta \to 0,
\end{equation}
where $\pi : \PP(V) \to \PP^n$ is the natural projection.
Since $\zeta^{(\frac{n(n+1)}{2}-1)}$ computes  the degree of the morphism $\psi$
 it is enough to
show that, for $n$  even,
$$\zeta^{(\frac{n(n+1)}{2}-1)}= \pi^*(h^n)\zeta^{\frac{n(n-1)}{2}-1}.$$

Since $V$ is vector bundle of rank $n(n-1)/2$ in the cohomology ring of
$\PP(V)$ the relation
\begin{equation}\label{eq10}
 \zeta^{\frac{n(n-1)}{2}} = \sum_{i=1}^{n}(-1)^{i+1}\pi^*(C_i(V))\zeta^{\frac{n(n+1)}{2}-i}
\end{equation}
holds, where $C_i(V),\, (1\leq i \leq n)$ are the Chern classes of $V.$
Repeated use of the relation \ref{eq10}  gives
\begin{equation}\label{eq11}
\zeta^{\frac{n(n+1)}{2}-1} = (-1)^n\pi^*(S_n(V))\zeta^{\frac{n(n+1)}{2}-1},
\end{equation}
where $S_n(V) $ is the $n$th Segre class of $V.$
By equation (\ref{eq7}) and the fact that the  total Segre class of $V$ is the
total Chern class  $C(T_{\PP^n}(-2))$ of $T_{\PP^n}(-2).$ Tensoring the equation (\ref{eq0}) by
$\OOO_{\PP^n}(-1)$ to obtain
\begin{equation}\label{eq12}
C(T_{\PP^n}(-2))= (1-h)^{n+1}(1-2h)^{-1},
\end{equation}
where $h$ is the class of the line bundle $\OOO_{\PP^n}(1)$ in ${\rm H}^2(\PP^n, \ZZ).$
From the equation (\ref{eq12}) we deduce that
$$C_n(T_{\PP^n}(-2)) = (\sum_{i=0}^n(-1)^i\binom{n+1}{i}2^{n-i})h^n$$
$$=((-1)^{n+1}\frac{(\sum_{i=0}^n\binom{n+1}{i}(-2)^{n+1-i})}{2})h^n$$
$$= ((-1)^{n+1}(\frac{(1-2)^{n+1}}{2})+(-1)^{n+2}\frac{1}{2})h^n.$$
Thus
$$S_n(V)= \left\{\begin{array}{lcr}
h^n & {\rm if} & n \,\,\, {\rm even}\\
0 & {\rm if} & n \, \,\,{\rm odd}
\end{array} \right.
$$
Now from equation (\ref{eq11}) we deduce that, for $n$ even
$$\zeta^{\frac{n(n+1)}{2}-1}= \pi^*(h^n)\zeta^{\frac{n(n-1)}{2}-1}.$$
Since $\pi^*(h^n)\zeta^{\frac{n(n-1)}{2}-1}$ is
the class of a point in $\PP(V)$ it follows that the degree of the map $\psi$ is one
when $n$ is even.
$\hfill{\Box}$

 {\bf Remark:} {\it The bi-rational morphism $\psi$  can be described geometrically
as follows:
   let
   $$ S = \{ (v, \varphi)\in \PP^n \times \PP(Alt({n+1}))| (\varphi)(v)=0 \}.$$
   For $v \in \CC^{n+1}\setminus \{0\}$ the the the linear subspace $\phi \in Alt({n+1})$
   gets identified with fibre  $\wedge^2(T_{\PP^n}(-1))_{v}$ of
   $\wedge^2(T_{\PP^n}(-1))_{v}$ over the point $[v] \in \PP^n$ hence
  $S \simeq \PP(V).$   Under this identification, when $n$ is even, the bi-rational morphism
  $\psi$ is the projection onto the second factor. From now on we assume $n$ is an even
  integer say $n=2k.$ Let
   $$D= \{\varphi \in \PP(Alt{n+1})|\,\, \text{rank}(\varphi)\leq n-1\},$$
   $F = \phi^{-1}(D)$ and $U= \PP(E)\setminus F.$
The morphism $\psi$ is one-one on the open set $U.$ }

Next we prove that the set $F$ in the previous Remark is a divisor and we identify
its class in the Picard group of $\PP(V).$ Note that Picard group of $\PP(V)$
is equal to ${\rm H}^2(\PP(E),\ZZ) = \ZZ[\pi^*(h)] \oplus \ZZ [\zeta].$

\begin{lem}\label{lem2}
	With the notations of previous Remark the set $F$ is a divisor in $\PP(V)$
	and its divisor class is $ k[\zeta]-[\pi^*(h)],$ where $k =n/2.$
\end{lem}

{\bf Proof:}  By equation (\ref{eq7})
we obtain a long exact sequence for $k$-th symmetric power of $V$
\begin{equation}\label{eq13}
0 \to \wedge^k(T_{\PP^n}(-1))\to \OOO_{\PP^n}^{\frac{n(n+1)}{2}} \otimes \wedge^{k-1}(T_{\PP^n}(-1)))\to \cdots
\end{equation}
\begin{equation}\label{eq14}
\cdots \to S^{k-1}(\OOO_{\PP^n}^{\frac{n(n+1)}{2}})\otimes (T_{\PP^n}(-1)) \to
S^{k}(\OOO_{\PP^n}^{\frac{n(n+1)}{2}}) \to S^k(V)\to 0.
\end{equation}
Tensoring the long exact sequence (\ref{eq13}) (\ref{eq14}) by $\OOO_{\PP^n}(-1)$
 and computing the cohomology
we deduce
$${\rm H}^0(\PP(V), \zeta^{\otimes k}\otimes \pi^*(\OOO_{\PP^n}(-1)))\simeq
{\rm H}^0(S^k(V)(-1)) $$
$$ \simeq {\rm H}^k(\wedge^k(T_{\PP^n}(-1))(-1)) \simeq
{\rm H}^k((\Omega_{\PP^n}^k)) \simeq \CC,$$
where the $\Omega_{\PP^n}^k$ is the bundle $k$ differential forms and the last isomorphism is the consequence of Bott's formula [See, Page 8 \cite{CMH}].
This proves that the divisor class $ k[\zeta]-[\pi^*(h)]$ contains a unique effective divisor.
From the exact sequence (\ref{eq7}) we deduce that the Segre class $S_{n-1}(V)$
is equal to $(-1)^{n-1}kh^{n-1}$ and hence
$$(k[\zeta]-[\pi^*(h)]).\zeta^{\frac{n(n+1)}{2}-2}= $$
$$\pi^*(h^n)\zeta^{\frac{n(n-1)}{2}-1}-(-1)^{n-1}\pi^*(h)\pi^*(S_{n-1}(V))\zeta^{\frac{n(n+1)}{2}-1}= 0.$$
For $v \in \PP^n$
the fibre  $V_v$ is isomorphic to ${\rm Alt}(\CC^n, {\CC^{n+1}/{\CC.v}}).$ For
$\varphi \in Alt{n+1}$  rank of $\varphi$ is less than or equal to $n-1$
if and only if image of $\bar{\varphi}\in {\rm Alt}(\CC^n, {\CC^{n+1}/{\CC.v}})$ of $\varphi$
has rank less than or equal to $n-1,$ i.e., $\bar\varphi$ is not an isomorphism.
 Since the complement of
isomorphisms in ${\rm Alt}(\CC^n, {\CC^{n+1}/{\CC.v}})$ is given by vanishing of the
homogeneous polynomial of degree in $k,$ i.e., section of $\OOO_{\PP(V_{v})}(k).$
The set
$$ F_v =\{\bar{\varphi} \in {\rm Alt}(\CC^n, {\CC^{n+1}/{\CC.v}})|  {\rm rank}(\bar\varphi) \leq n-1 \} $$
is an irreducible divisor given by the vanishing of the restriction to  $\PP(V_v)$ of the
non-zero section (unique upto non zero scalar) of
$${\rm H}^0(\PP(V), \zeta^{\otimes k}\otimes \pi^*(\OOO_{\PP^n}(-1)))$$
The non-zero section $t$ (unique upto multiplication by a
non zero scalar) determines a section of $\OOO_{\PP(V_{v})}(k),$ namely  Pfaffian
whose square is the determinant of skew symmetric form. It is clear that $F_v$ is the
set of zeros of  the section $t|_{\PP(V_v)}.$  This proves that the set $F$ is the support of the
divisor the $k[\zeta]-[\pi^*(h)].$             $\hfill{\Box}$

{\bf Remark 2:} {\it Theorem (\ref{thm2}) can be used to obtain  for
blow up of $\PP^m, \,(2n-1 \leq m \leq \frac{n(n-1)}{2}-1)$   along a non linear sub-variety a projective
bundle structure over $\PP^n,$ for even integer $n \geq 4.$}

{\bf Remark 3:} The special case of Theorem (\ref{thm2}) appears first in \cite{EL} and has
been used in the context of Quantum Cohomology in \cite{CCGK}, \cite{CGKS}, \cite{AS1},
\cite{AS2}, \cite{AS3}.

{\bf Acknowledgement: } We would  like to thank organisers of the  conference
on "Derived Category and Algebraic Geometry" at TIFR Mumbai for the invitation to give a talk and TIFR for its hospitality.

\providecommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{arXiv:#1}}
\providecommand{\doi}[1]{\href{http://doi.org/#1}{\tt doi:#1}}


TITLE: Non-existence of gluon singlets: Any recent theoretical progress?
QUESTION [11 upvotes]: An unanswered question from last year (2012) on gluon singlets asked whether there is any theoretical explanation for the experimental absence of the ninth or colorless (singlet) gluon. This is the gluon that, if it existed, would give allow the strong force to extend far beyond the range of atomic nuclei, with catastrophic results. (Non-experts can find an explanation of the singlet gluon in the Addendum below.)
Qmechanic partially answered the question with a link to this 1996 paper by J.J. Lodder, which proposed that the singlet gluon exists but is so massive that its effects are negligible.
However, Lubber's massive-singlet idea seems to have gone nowhere. On an cursory search I could not find any references to the 1996 draft, even though it was written seventeen years. I also do not find Ludder's approach very plausible. Theoretically he had to do a bit of mayhem to Standard Model symmetries to come up with his model. More importantly, though, its seems unlikely that the idea holds water experimentally. The mass of the singlet gluon would have to be astronomical indeed for it not to have shown up in high-energy experimental results, especially in the post-Higgs era.
So, my version of the singlet question is this: Do there exist any plausible, Standard Model compatible theories for the observed absence of the singlet gluon, other than the apparently non-starter idea that the singlet exists but is very massive? Or alternatively, have searches for the Higgs boson produced any evidence that the singlet gluon state may in fact exist and have a very large mass?

Addendum for non-experts: What is a "gluon singlet" and why is it important?
One way to understand the "colorless" gluon singlet is through an analogy with how photons work for the electric force.
Photons interact with electrically charged electrons, but do not carry any electric charge themselves. Because they have no charge, even low-energy photons can easily escape an atom or electron and travel infinite distances. Imagine, however, what would happen if photons did carry electric charge. Such photons would have the same energetic difficulties leaving a neutral atom as an electron, dramatically altering and limiting how they behave. (I should probably mention that electrically-charged photons in one sense really do exist: They are more-or-less the $W^\pm$ particles of electroweak theory.)
Curiously, the charged situation is reversed for the otherwise photon-like gluons. It is the gluons that convey the strong force, and thereby hold quarks together to form protons and neutrons, as well as secondarily binding protons and neutrons together within atomic nuclei.
There are eight types of gluons instead of just one, due to there being more than two types of charge in the strong force. However, in sharp contrast to photons, all of the eight gluons normally carry strong (color) charge. The color charges of gluons cause them to interact strongly with the quarks that emit them and with each other. As in the earlier hypothetical example of how electrical charges would dramatically limit the range of photons, the presence of color charges on gluons similarly limits the distances over gluons can convey the strong force. Consequently, the strong force has almost no impact beyond the scale of atomic nuclei.
However, the same mathematical model that predicts the eight color-charged gluons also predicts a ninth neutral or strong-charge-free gluon, called the singlet gluon, that has never been seen experimentally. Its lack of color would make its impact far greater than that of any of the eight other gluons. In particular, just as charge-free photons can carry the electric force far beyond the range of atoms, a charge-free gluon, if it existed, would allow the strong force to extend far beyond the range of nuclei.
The repercussions would be huge. In fact, the non-existence of the singlet gluon is best demonstrated by the fact that we exist at all. My best guess (only that, since I have not seen any papers on it) is that if the colorless singlet quark really did exist, every clump of two or more atoms in the universe would melt together into an amorphous sea of quarks. Even if that is not correct, I assure you it that the consequences of the existence of singlet gluons would be... very bad indeed!
That is also why I find I think the non-existence of the singlet gluon is a truly interesting question, one that probably deserves more theoretical attention than it has received over the decades since strong force theory was first codified.

REPLY [7 votes]: For there to be a color singlet gluon color theory would have to be a U(3) theory, not SU(3), but the great weight of experimental evidence assembled over many years supports SU(3) not U(3).  In the 60's (before the Standard Model had reached maturity) a couple papers by well-known theorists appeared investigating the possibility of U(3) rather than SU(3) symmetry, but that's about it.  One of the papers worked out a scheme for integer rather than fractional charge for quarks, but none of the main ideas ended up in the Std Model, which is still firmly SU(3) => no singlet gluons.  In addition there is no experimental evidence for any long range component to the strong force - quite the opposite.   There are plenty of unaswered questions in physics and I suspect that most theorists feel that (if they even give the ninth gluon a moment's consideration at all  :-)  ) there are many other topics to research that are more likely to lead to interesting new results.


TITLE: Moment of Inertia of a solid hemisphere. What am I doing wrong?
QUESTION [0 upvotes]: I want to calculate the MOI of a uniform solid hemisphere about

Axis passing through its centre of mass (COM) and perpendicular to the circular base.

Axis coinciding with any diameter at the circular base.

To proceed with this, I am using the fact that a solid sphere of mass $M$ and radius $R$ has $$I=\frac{2MR^2}{5}$$
Hence by symmetry argument I am getting the same Moment of Inertia in both the above 2 cases which I think is incorrect.

REPLY [1 votes]: Your symmetry argument is correct, this is because say we have full sphere and an axis passing through its center. Breaking it into 2 symmetrical hemispheres means each carries half of the total moment of inertia that is $\frac{1}{5}MR^2$. But its mass is also half of total so its moment of inertia is $\frac{2}{5}MR^2$, as long as it passes through its center of the total sphere.


TITLE: Find the number of Diagonal Matrices with product of diagonal elements as $27000$
QUESTION [0 upvotes]: Find the number of $3 \times 3$ Diagonal Matrices with product of diagonal elements as $27000$ with diagonal elements as Natural numbers.
Let $x_1$,$x_2$ and $x_3$ are diagonal elements then we have to find all ordered triples $(x_1,x_2,x_3)$ such that $$x_1 \times x_2 \times x_3=27000=2^3 \times 3^3 \times 5^3$$
Case $1.$ The number of Matrices with two diagonal elements as unity is obviously $3$
Case $2.$ The number of matrices with exactly one diagonal element as unity is given by $$ \binom{3}{1} \times \left((3+1)(3+1)(3+1)-1\right)=189$$
But how can we proceed if None of the Diagonal elements is Unity?

REPLY [1 votes]: I would attack it in a different manner.  If we just think about the factors of $2$ we need to find the number of weak compositions of $3$ into $3$ parts.  The stars and bars argument in Wikipedia shows there are ${5 \choose 2}=10$ of them.  We can then multiply the number of choices for each prime, getting $10^3=1000$ ordered factorizations of $27000$