text:
\section{Poincar\'e series and Euler characteristic of supermaximal models

}
\label{sec:poincaresupermodels}
In this section we use  the cohomology basis described in Theorem \ref{thm:cohomology} to prove   a formula for a series that encodes all the information regarding the Poincar\'e polynomials of the supermaximal models \(Y_{\Bc(n-1)}\).

We start by recalling the analogue computation in the case of minimal models.
The  Poincar\'e series \[\Phi(q,t)=t+ \sum_{n\geq 2}\sum_i dim \ H^{2i}(Y_{minA_{n-1}}, \Z) q^i \frac{t^n}{n!}\] for the minimal De Concini-Procesi models \(Y_{minA_{n-1}}\)  has been computed in many different ways.
One can see for instance  \cite{Manin}, \cite{YuzBasi}, \cite{GaiffiBlowups},  \cite{getzler};  a formula for another series that encodes the same information   is provided   in Section \ref{ss:action3} of the present paper.

Both in   Section 5 of \cite{YuzBasi} and  Section 4 of \cite{GaiffiBlowups} the computation of \(\Phi(q,t)\) consists in counting  the elements of  the Yuzvinski basis for \(H^{*}(Y_{minA_{n-1}}, \Z)\) described in Section \ref{constru}.
We recall that, given  a monomial \(m_f\) in this basis, the set \(supp \ f\) is a nested set and that in  Section \ref{secexamplean} we represented the $\Fc_{A_{n-1}}$- nested sets, i.e. the elements of \(\Nc( \Fc_{A_{n-1}}), \) as oriented forests on \(n\) leaves. Let us then denote by \(\lambda(q,t)\)  the contribution  to \(\Phi(q,t)\)  provided by the basis monomials whose associated nested set is represented by a tree.

It turns out that \(\lambda(q,t)\)  satisfies the following recursive relation:
\begin{displaymath}
	\lambda(q,t) ^{(1)}=1 + \frac{\lambda(q,t) ^{(1)}}{q-1}
	\left[e^{q\lambda(q,t) }-qe^{\lambda(q,t) }+q-1    \right]
\end{displaymath}
(here the superscript \(^{(1)}\) means the first derivative with respect to \(t\)).
Then one obtains a formula for   \(\Phi(q,t)\) by observing that   \(\Phi(q,t)= e^{\lambda(q,t) }-1\).

Let us now denote by \(\Phi_{super}(q,t)\) the  Poincar\'e series:
\[\Phi_{super}(q,t)=t+ \sum_{n\geq 2} \sum_i dim \ H^{2i}(Y_{\Bc(n-1)}, \Z) q^i \frac{t^n}{n!}\]
\begin{definition}
We define the following series in four variables:
\begin{equation}
\label{formulacsi}
 \xi(t,q,y,z)=\Phi(q,t) + \sum_{
\begin{array}{c}
 n \geq 2  \\
\MS\in \Nc'( \Fc_{A_{n-1}}) \\
 |\MS|=\ell +1 > 1
\end{array}}\sum_{r\geq 0, k\geq 0}dim \ H^k(\Dc_\MS)N_{r,\MS} \ y^\ell z^r q^k\frac{t^n}{n!}
\end{equation}
where   \(\Dc_\MS\) is the subvariety in the boundary of  \(Y_{min A_{n-1}}\) defined in Section \ref{constru}

and  \(N_{r,\MS}\) is the number of nested sets in \(\Nc'( \Fc_{A_{n-1}})\)   that contain \(\MS\) and whose cardinality is \(|\MS|+ r\).

\end{definition}
\begin{theorem}
\label{teo:sostituzioni}
One can obtain  the Poincar\'e series \(\Phi_{super}(q,t)\)  from the series  \(\xi(t,q,y,z)\)  by substituting:
\begin{itemize}

\item  \(y^\ell\)  with \(\frac{q^\ell-q}{q-1}\);

\item  \(z^r\)   with \[\sum_{s\leq r, \;  j_0=0< j_1< \cdots  < j_s=r} \frac{r!}{j_1!(j_2-j_1)!\cdots (j_s-j_{s-1})!}\prod_{\theta=1}^{s}q\frac{q^{j_{\theta}-j_{\theta-1}-1}-1}{q-1}\]

\end{itemize}
\end{theorem}
\begin{proof}
The monomials in the basis of   \(H^*(Y_{\Bc(n-1)})\) (\(n\geq 2\)) are described by  Theorem \ref{thm:cohomology}:
\[  \eta\ c_{\MS_1}^{\delta_1}c_{\MS_2}^{\delta_2}\cdots c_{\MS_k}^{\delta_k}  \]
where \(\eta\) is  represented by a monomial in the basis of  \( H^*(\Dc_{\MS_1})\) if \(k\geq 1\) and of \( H^*(Y_{min A_{n-1}})\) if \(k=0\).
The monomials with \(k=0\) are computed by the series \(\Phi(q,t)\) that is the first addendum in the formula (\ref{formulacsi}).
Then we observe that, once \(\MS_1\) is fixed, if  the  exponent of the variable \(z\) is \(r\) this means that we are keeping into account all the monomials such that \(|\MS_k-\MS_1|=r\). This is expressed by the substitution formula  for \(z^r\).

The exponent of the variable \(y\) coincides with \(|\MS_1-\{V\}|\) and  the exponent  \(\delta_1\) satisfies \(1\leq \delta_1\leq |S_1-\{V\}|-1\): this is expressed by the substitution of \(y^\ell\)  with \(\frac{q^\ell-q}{q-1}\) (notice that \(\ell>1\)).

\end{proof}

We are therefore interested in  finding a formula for \(\xi(t,q,y,z)\).
Let us denote by  \(p_n(q)\), for \(n\geq 2\),  the Poincar\'e polynomial of the minimal model \(Y_{minA_{n-1}}\), i.e. we can write    \(\Phi(q,t)=t+ \sum_{n\geq 2}p_n(q)\frac{t^n}{n!}\).
Then we consider the series  \(W(t,z)=\sum_{n\geq 2}w_n(z) t^n\)  where the polynomials \(w_n(z)\) count the number of elements of $\Bc(n-1)$. More precisely  \[w_n(z)=\sum_{1\leq j\leq n-1}\sum_{
\begin{array}{c}
  \MS\in \Bc(n-1)    \\
  |\MS|=j
\end{array}
}z^{j-1}\]
As we will  recall in Section \ref{sec:combinatorialaction},
we have
\[w_n(z)=\sum_{1\leq j\leq n-1}|\Pc_2(n+j-1,j)|z^{j-1}\]
where we  denote by \(\Pc_2(n+j-1,j)\)  the number of unordered partitions of the set \(\{1,..., n+j-1\}\) into \(j\) parts of  cardinality greater than or equal to 2 (these numbers are  the 2-associated Stirling numbers of the second kind, see  for instance the table at  page 222 of \cite{comtet}).

If we now put  \[\psi(t,q,z)=\sum_{n\geq 2}p_n(q)w_n(z)\frac{t^n}{n!}      \]
we can write a formula that computes  the series \(\xi(t,q,y,z)\):
\begin{theorem}
\label{teo:formulapoincare}
We have
\[  \xi(t,q,y,z)= \Phi(q,t)+ \sum_{\nu\geq 1}  \psi(t,q,z)^{(\nu)}  \frac{\Gamma^\nu}{\nu !} \]
where
\[\Gamma=\sum_{\ell\geq 1}\frac{1}{\ell !}y^\ell (\psi(t,q,z)^\ell)^{(\ell-1)}\]
and  the superscript \(^{(j)}\) means the \(j\)-th derivative with respect to \(t\).
\end{theorem}
The next (sub)section will be devoted to the proof of this theorem.
\begin{example}
Here it is a computation of the first terms of the series  \(\xi(t,q,y,z)\), that is the terms whose \(t\)-degree is \(\leq 5\):

$${\frac{t^5}{5!}}\left[\left(315\,q^2+1305\,q+315\right)yz^2+\left(\left(
 315\,q+315\right)y^2+\left(210\,q^2+870\,q+210\right)y\right)z
 \,+\right.
$$
$$
\left.+\,105\,y^3+\left(105\,q+105\right)\,y^2+\left(25\,q^2+95\,q+25\right)
 \,y+q^3+16\,q^2+16\,q+1\right]\,+
$$
$$
+\frac{t^4}{4!}\left[30yz(1+q)+15 y^2+10y(1+q) +q^2+5q+1\right] +\frac{t^3}{3!}\left[3y+q+1\right] +\frac{t^2}{2}+t.$$
\end{example}

If we put \(q=1\) in the formula for the Poicar\'e series \(\Phi_{super}(q,t)\), we obtain the Euler characteristic series of  the supermaximal models \(Y_{\Bc(n-1)}\). Moreover we notice that if we put \(q=-1\) we obtain the Euler characteristic series of the {\em real} supermaximal models \(Y_{\Bc(n-1)}(\R)\) constructed with base field \(\R\).
In fact from a result of  \cite{krasnov} it follows   that
\(H^{2i}(Y_{\Bc(n-1)}, \Z_2)\cong H^{i}(Y_{\Bc(n-1)}(\R), \Z_2)\);  therefore \(\sum_{i}(-1)^i \dim\ H^{2i}(Y_{\Bc(n-1)}, \Q)\)
 is equal to the Euler characteristic  \(\chi_E (Y_{\Bc(n-1)}(\R))\).

We observe that one can obtain  the Euler characteristic series in a more direct way  from the series \(\xi(t,q,y,z)\):

\begin{theorem}
\label{teo:eulerchar}
One can compute the Euler characteristic series of  the models \(Y_{\Bc(n-1)}(\R)\) from  the series \(\xi(t,q,y,z)\)
by  substituting  \(q\) with \(-1\) and also
\begin{itemize}

\item \(y^k\) with \(-1\) if \(k\) is even, otherwise with 0;

\item \(z^r\)  with \(E_r\), where we denote by \(E_n\)  the Euler secant number defined by
\[\frac{2}{e^t+e^{-t}}=\sum_nE_n\frac{t^n}{n!}\] (see sequence A028296 in OEIS; notice that if \(n\) is odd then \(E_n=0\)).

\end{itemize}
\end{theorem}
\begin{proof}

When we find \(z^r\) in \(\xi(t,q,y,z)\), it means that we are computing the contribution to the Euler characteristic of all the  factors
\( c_{\MS_1}^{\delta_1}c_{\MS_2}^{\delta_2}\cdots c_{\MS_k}^{\delta_k}\) of the basis  elements,   such that \(|\MS_k-\MS_1|=r\).
We observe that this contribution is non zero if and only if \(r\) is even. In this case our problem is equivalent to  computing the Euler characteristic of the order  complex   of the poset \(\Pc(\{1,...,r\})_{even}\) of the subsets of \(\{1,...,r\}\) with even cardinality.
By  Philip Hall Theorem (see for instance Proposition 1.2.6  of \cite{wachsposet})  we can compute it via the Moebius function, that  is equal to \(E_r\)  (see  Section 3.7 of Stanley's  survey   \cite{stanleysurvey}).

\end{proof}

\subsection{Proof of Theorem \ref{teo:formulapoincare}}
We will follow a strategy similar to the one in Section 3 of \cite{Gaiffigeneralizedpoincare}.

 \begin{definition}
Given any rooted oriented  tree $T$, the polynomial  ${\cal Q}( T)$
is a product of
monomials that contains a factor for each vertex  of $T$: if $v$
is a vertex of $T$ with  $\nu $ outgoing edges, the
corresponding factor is $y^\nu{\displaystyle \psi(t,q,z)^{(\nu )}}$.

\end{definition}

We notice that  the cardinality
${\displaystyle Aut (T)}$  of the automorphism group of a rooted
tree  $T$  is equal to  the product ${\displaystyle \prod
_{v}^{}\gamma _{sym,v}}$ where $v$ ranges over the vertices of $T$ and $\gamma _{sym,v}$ is determined in this way:  delete $v$ and consider the
connected components of the subgraph of $T$ that stems from
$v$. Suppose that they can be partitioned  in $k$ automorphism classes with
the following cardinalities: $a_1,a_2,\ldots,a_k$. Then $\gamma
_{sym,v}=a_1!a_2!\cdots a_k!$.

Now  we  associate a tree  to every monomial \( \eta\ c_{\MS_1}^{\delta_1}c_{\MS_2}^{\delta_2}\cdots c_{\MS_k}^{\delta_k}\) in the basis of \(H^*(Y_{\Bc(n-1)})\),  by choosing the rooted tree  with \(n\) leaves associated with the nested set \(\MS_1\) (see Section
\ref{secexamplean}) and deleting the leaves and the edges that contain a leaf.
Then \(\xi(t,q,y,z)- \Phi(q,t)\) can be computed by regrouping together all the basis monomials that are associated with the same rooted tree. This means that \(\xi(t,q,y,z)- \Phi(q,t)\) can be written  as:

\begin{displaymath}

	\sum _{n\geq 1}^{}\psi(t,q,z)^{(n)}\frac{(y\Gamma)^n}{n!}
\end{displaymath}
where
\begin{displaymath}
	\Gamma=\sum
	_{[T] }
	\frac{{\cal Q}( T)}{Aut (T)}
\end{displaymath}
and the sum ranges over all the automorphism classes of  nonempty oriented,
rooted trees.

\noindent Thus the problem can be reduced to the one of finding a `nice' formula for
$\Gamma$. This is provided by the following theorem, that is a consequence of Theorem 3.3   of \cite{Gaiffigeneralizedpoincare}.

\begin{theorem}
\begin{equation}
	\sum
	_{
	\begin{array}{c}
	[T]
	\end{array}
	}^{}\frac{{\cal Q}( T)}{Aut (T)}=
	\sum _{n\geq 1}^{}\frac{1}{n!}y^{n-1}\left(\psi(t,q,z)
	^n \right)^{(n-1)}
	\label{bellaeq}
\end{equation}
where $[T]$ ranges over all the automorphism classes of
nonempty oriented  rooted trees.
\label{belte}
\end{theorem}


TITLE: Continuity between normed linear spaces, argument verification
QUESTION [1 upvotes]: Let $V,W$ be any finite dimensional normed real vector spaces, and $X$ any normed real vector space. Let $B:V\times W \to X$ be a bilinera map. I want to show that $B$ is continuous.
My argument
Since on a finite dimensional real vector space all norms are equivalent, I can choose to consider the following norms on $V,W$ and $V\times W$ : choose a basis $(E_1,\dots, E_n)$ of $V$ and $(F_1,\dots,F_m)$ of $W$ and define $$|\cdot|_V:V\to [0,+\infty), v=\sum_i v^iE_i\mapsto\sum_i|v_i|$$ and $$|\cdot|_W:W\to [0,+\infty), v=\sum_i v^iF_i\mapsto\sum_i|v_i|$$, and $$|\cdot|:V\times W\to [0,+\infty), (v,w)\mapsto |v|_V+|w|_W$$
Now we have $B(v,w)=\sum_{i,j}v^iw^jB(E_i,F_j)$ and putting $M=\max\{|B(E_i,F_j)|_X:i,j\}$ we have
$$|B(v,w|\leq M\sum_{i,j}|v^iw^j|\leq Mnm |v||w|$$ and from this it follows
$|B(v,w)-B(h,k)|=|B(v,w)-B(v,k)+B(v,k)-B(h,k)|\leq|B(v,w-k)|+|B(v-h,k)|\leq Mmn[|v||w-k|+|v-h||k-w|+|v-h||w|]\leq Mmn [|v|+|(v,w)-(h,k)|+|w|]|(v,w)-(h,k)|$.

Is my argument correct? I need this for showing that $B$ is differentiable according to the definition of Lee's "Introduction to smooth manifolds " 2° Edition Appendix C pag 642.

In my proof of differentiability I need continuity so I'm proving this first.

If you know shorter ways of showing continuity of $B$ let me know. Also let me know if you can prove differentiability of $B$ without first showing continuity.

N.B. I can only use definitions and results previously showed in that book.

REPLY [1 votes]: Your arguments for continuity are correct.
If you know about tensor product, you can make an alternative proof:
Observe that $B:V\times W\to X$ is just the composition of the following maps:
$$V\times W\ \overset{\vartheta}\to\ V\otimes W\ \overset{\hat B}\to\ S\ \hookrightarrow\ X$$
where $\vartheta$ is the canonical bilinear map $(v,w)\mapsto v\otimes w$; $\ \hat B$ is the linear map induced by $B$, mapping $v\otimes w\mapsto B(v,w)$, and $S:={\rm ran}(\hat B)$ is its image, which is included in $X$.
Now, all these, except perhaps $X$, are finite dimensional real vector spaces, so both $\vartheta$, as a bilinear map, and $\hat B$, as a linear map, are differentiable (even smooth).
So is the inclusion of $S$ into $X$, as that's also a bounded linear map.

Edit: Actually, we can even skip the tensor product part, and define $S$ as the range of the (bilinear) function $B$, and show that it's a finite dimensional subspace (its dimension is $\le\dim V\cdot\dim W$).


TITLE: Show that there is no affine transformation that takes a circle to a hyperbola in $\mathbb{R}^2$
QUESTION [7 upvotes]: "Show that the standard circle (defined by $f(x,y) = x^2 + y^2 - 1$) is not equivalent to the standard hyperbola (defined by $g(x,y) = x^2 - y^2 - 1$). That is, show that there is no $[A,\overline{s}] \in \text{Aff}(\mathbb{R}^2)$ such that $[A,\overline{s}] \cdot f(x,y) = g(x,y)$. Check that there is such an $[A,\overline{s}]$ if we allow $A \in \text{GL}_2(\mathbb{C}).$"
I've reduced this to showing that there are no $a,b,c,d,s,t \in \mathbb{R}$ such that $$f(ax+as+by+bt,\: cx + cs + dy+dt) = g(x,y).$$
$$\Rightarrow(ax+as+by+bt)^2+(cx + cs + dy+dt)^2 - 1=x^2-y^2-1$$
How should I proceed? Expanding that expression probably isn't the best way to do it.

REPLY [5 votes]: Why think when you can simply compute?
So $f(x,y)=x^2+y^2-1$. Consider the affine transformation given by $$\left\{\begin{array}{l}x\leftarrow ax+by+s\\y\leftarrow cx+dy+t\end{array}\right.$$ Applying it to $f$ we get $$f(ax+by+s, cx+dy+t)=(a^2+c^2) x^2+2(a b + c d)xy+(b^2 + d^2)y^2+\text{terms of lower degree}.$$ If this is to be equal to $g(x,y)=x^2-y^2-1$, then, in particular, looking at the coefficient of $y^2$ we see that we must have $$b^2+d^2=-1.$$ Of course, this is not going to work...

REPLY [2 votes]: Let us call $f_0$, $f_1$ and $f_2$ the parts of $f$ which are homogeneous of degree $0$, $1$ and $2$, respectively, so that in particular $f=f_2+f_1+f_0$, and similarly for $g$.
If $[A,s]$ is an invertible affine transformation and $[A,s]\cdot f=g$, then one checks easily that $[A,s]\cdot f_i=g_i$ for each $i\in\{0,1,2\}$. In particular, $$[A,s]\cdot(x^2+y^2)=x^2-y^2.$$
But two polynomials which are affinely equivalent are either both irreducible or both reducible, yet $x^2+y^2$ is irreducible and $x^2-y^2$ is not.
This argument is «geometric»: we are using the fact that a circle has no asymptotic directions while the hyperbola has two, and that the property of having asymtotic directions is preserved under affine equivalence.


TITLE: What are the major difficulties to the production of elements?
QUESTION [0 upvotes]: Searching online about this topic only gives results about nucleosynthesis (making elements by nature) and the 24 radioactive artificial elements.
Why isn't there any information about methods to make any common stable elements, especially about making rare earth elements, from abundant elements like hydrogen/silicon?
Mass production of elements may be not cost effective than nature resource for now (who will know what breakthrough can make it a different story in the future?), but at least for curiosity any common elements could be tried to be made in the labs, with something like super high energy laser which can even emulate supernova environments.
It's like:
Nature law+Invention by human=Anything
Are they already made, or what is the reason that they (or some of them) can't be made? What is the difficulty towards the invention of such methods and apparatus?

REPLY [2 votes]: To get the information you wish, you'll need to dig a little deeper, as follows.
For any element in the periodic table, it is possible to calculate the quantity of work required to squeeze its constituent protons and neutrons together hard enough to stick. This is called the nuclear binding energy and is analogous to the heat of formation of a chemical compound. How to calculate this?
Knowing the individual masses of the protons, neutrons and electrons in any element, you can sum these together and compare the sum with the actual atomic mass of the element. For elements lighter than iron, there is a net energy release when the protons and neutrons fuse together and the actual mass is less than the sum. That mass difference is called the mass defect which is equal (via E = mc^2) to the energy released when the protons and neutrons fuse themselves together. For elements heavier than iron, the actual mass is greater than the sum which means net positive work must be done to create that nucleus (i.e., the nucleus contains stored potential energy, again equal via E=mc^2 to the mass difference).
In your example, where you wish to manufacture rare elements by squeezing other common elements together, E= (delta m)c^2 is the number you want. However, note the following:
This process can be carried out in a particle accelerator, where lighter particles are shot at heavier nuclei to produce still-heavier nuclei. But this is hugely inefficient because 1) the energy cost to speed up the "bullets" is enormous and 2) the vast majority of the lighter bullets either miss their target nuclei entirely, bounce off without sticking, or actually shatter the target to pieces.
That gross inefficiency makes it completely infeasible to build up rare elements in the lab compared to the cost of digging holes in the earth's crust, hauling ore to the surface, and purifying it to get those same elements.


TITLE: Prove that V is a subset of $(V^{\perp})^{\perp}$
QUESTION [0 upvotes]: How can I prove that $V ⊆ (V^{\perp})^{\perp}$ for a subspace V of $\mathbb R^n$?
It makes sense intuitively, but I don't know how to do it mathematically. If anyone could give me a hint on how to start, that'd be great.

REPLY [3 votes]: Step 1: Let $v\in V$ be a vector.
Step 2: Set $U=V^\perp$, and prove that $v\in U^\perp$.  You will use the definition of $\perp$ repeatedly.
Step 3: Since $v\in V$ was arbitrary, this proves $V\subseteq U^\perp=(V^\perp)^\perp$.


TITLE: Show that the given collection S of subsets of the set X is a partition of X.
QUESTION [0 upvotes]: I feel as if I'm going in circles with this proof.

REPLY [2 votes]: Since I've offered a bounty and my reputation fell below 50, I'll simply leave a hint here instead of commenting, sorry!
Hint: a family is a partition iff it is a family of abstraction classes of a certain equivalence relation. And (under axiom of choice) Relation $R$ is equivalence iff there exists $f$ such that $aRb$ iff $f(a)=f(b)$. The formulation of a task involves a certain promising $f:Z^+ \times Z^+ \rightarrow Z^+$...


TITLE: Accumulated value problem - Find X using pthly annuity
QUESTION [1 upvotes]: A family wishes to accumulate 25,000 in a college education fund at the end
of 10 years. If they deposit 750 in the fund at the end of every six months for the first 5
years and 750 + X in the fund at the end of every six months for the second 5 years, find
X if the fund earns at effective interest rate 8%. (You must use the pthly annuity method
(ie. $\ a_{\overline {n}\rceil}$ or $\ s_{\overline {n} \rceil}$ to solve this question)
This is what I have come up with so far and I am not getting the correct answer ($X=237.72$ is the book's answer)
$25,000 = 750\ s_{\overline {5} \rceil}^{(2)} *(1+8\%)^5 + (750+x)\ s_{\overline {5} \rceil}^{(2)}$

REPLY [1 votes]: You have to transform the yearly interest rate $i$ into an equivalent semi-annual interest rate $i^{(2)}$.
$1+i=q=1.08\Rightarrow \sqrt{1+i}-1=\sqrt{1.08}-1=i^{(2)}$
And $(1+i^{(2)})^{m}=(1+i)^{m/2}$
Therefore the equation is
$$750\cdot 1.08^5\cdot \frac{1.08^5-1}{1.08^{0.5}-1}+(x+750)\cdot \frac{1.08^5-1}{1.08^{0.5}-1}=25,000$$


TITLE: Negate the following statements formally so that **no negation symbol remains:**
QUESTION [1 upvotes]: Negate the following statements formally so that no negation symbol remains:
(i)  $ \ \ \forall y \exists x (y>0 \to x \leq 0 ) \ $
(ii) $ \ \forall x \forall y \exists z(x <z \leq y ) \ $
Answer:
My approach is as follows:
(i)
The negation of the statements in (i) without negation symbol is
$ \exists y  \ \ s.t. \ \  \forall x (x>0  \to y \leq  0) \ $
(ii)
Given
$ \forall x \forall y \ \exists z \ ( x <z  \leq y ) \  \\\sim \forall x \forall y \ \exists z \ ( x <z  \wedge z \leq y ) $
The negation is given as
$ \exists x \exists y \ \ s.t. \ \ \forall z  \ ( x \geq z \wedge  z > y ) $
Am I true ?
Is there any help ?

REPLY [0 votes]: You have the quantifier duality okay, and the negations for the order comparators, but not the negation of the predicate's connectives.

(i)  $\forall y\ \exists x\ (y>0\ \to\ x \leq 0 )$

Use $\neg(P\to Q)\iff (P\wedge\neg Q)$
(A conditional is falsified exactly when the antecedant is true but the consequent is false.)

(ii) $ \forall x\ \forall y\ \exists z\ (x <z\ \wedge\ z \leq y ) $

Use $\neg(P\wedge Q)\iff (\neg P\vee\neg Q)$
(deMorgan's Law of Connective Duality)


TITLE: Diffeomorphism invariance in special relativity
QUESTION [1 upvotes]: Suppose space time is the manifold $M $ isomorphic $ \mathbb{R^4}$ whit the metric $-\eta_{00}=\eta_{11}=\eta_{22}=\eta_{33}=1$ in the Cartesian coordinates $\Psi(p)=(x^0,x^1,x^2,x^3)$ for $p \in M $ .
So we have model $A=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $ for general relativity and a model $B=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $
for special relativity  where $\phi$ is a scalar field and $\frac{\partial}{\partial x^a }$ vector field.
It is often said that general relativity is diffeomorphism invariant and special relativity not.
For example, suppose we  have a diffeomorphism $F:M\rightarrow M$ and the wave
equation $$\eta^{ab}\frac{\partial \phi}{\partial x^a }\frac{\partial \phi}{\partial x^b }=0$$
We have by pullback $$F^*\left(\frac{\partial }{\partial x^a }\right)=\frac{\partial G^{b}}{\partial x^a }\frac{\partial}{\partial x^b },$$
$$F^*\phi=\phi \circ F$$
and $$F^*(\eta_{ab})=\frac{\partial F^{k}}{\partial x^a }\frac{\partial F^{j}}{\partial x^b }\eta_{kj}$$
where $G=F^{-1}$
So
$$F^*\left(\eta^{ab}\frac{\partial \phi }{\partial x^a }\frac{\partial \phi}{\partial x^b }\right)=\eta^{ab}\frac{\partial (\phi \circ F)}{\partial x^a }\frac{\partial (\phi \circ F)}{\partial x^b}$$
that is $\phi \circ F$ is a solution if is a solution $\phi$.
Is the reason that this does not work in special relativity is that in special relativity  the metric $\eta$ has to be fix.
That is  $$F^*B=\bigg \langle {M,F^*\eta,F^*\phi,F^*\phi\frac{\partial}{\partial x^a }} \bigg \rangle $$ is not a model of special relativity because $ F^*\eta \neq \eta$, and in general  relativity if $A$ is model than $C=\bigg \langle {M,g,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $ is a different model than $A$, but a model of general relativity.
What i mean is,though $F^*B$ and $B$ have equivalent solution they are not equivalent models because $F^*B$ is not a model of special relativity.
Is this the reason why special relativity is not diffeomorphism invariant?

REPLY [4 votes]: It is indeed "often said that general relativity is diffeomorphism invariant and special relativity is not", which is unfortunate because that's not true. Special relativity is perfectly diffeomorphism invariant, as is Newtonian mechanics and every other well-formulated physical theory. That's because diffeomorphisms (or, more precisely, isometries under which the metric tensor and all other fields on the manifold are appropriately pulled back and pushed forward along the diffeomorphism) simply represent changes of coordinates, and by definition the description of any truly physical quantity is covariant under changes of coordinates.
The only possible sense in which SR is any "less" diffeomorphism invariant than GR is in that Minkowski space happens to admit a natural family of global coordinate systems (Lorentz coordinates) in which the metric happens to take a particularly simple form, which is identically equal to $\pm \mathrm{diag}(-1, 1, 1, 1)$ - so simple that you can completely hide it into the raised and lowered indices, which have a trivial mathematical relation (just flippig the sign of the timelike component). In these coordinates, all of the covariant derivatives reduce to partial derivatives, so for elementary applications it's often easiest to always work in those coordinates and not worry about coordinate transformations. But it's still nothing more than just a particularly convenient choice of coordinates, and you're always free to switch to other coordinates. (Of course, the Riemann curvature tensor will vanish in any system coordinates.) In GR, there's no global choice of coordinates in general, and certainly not one as simple as Lorentz coordinates, so we're forced to deal with all the subtleties of changes of coordinates that we can get away with ignoring in SR.


TITLE: Showing a division justification?
QUESTION [3 upvotes]: I know this question may seem to trivial to discuss but here goes.
Let A,B,C be positive integers.
Propostion
If A divides B then A divides BC.
Explanation
Let
$B=AK$ K is integer
Consquently
$BC=(AK C)$
Anything that has A as a factor. Can be divided by A.
Therefore A divides BC.

REPLY [2 votes]: Yes, that's basically correct. I might be a little more explicit at the end: since $BC = AKC = A(KC)$ and both $K$ and $C$ are integers, so is $KC$; therefore $A$ divides $BC$.


TITLE: Volume of solid bounded
QUESTION [0 upvotes]: I'm trying to find the volume of the solid bounded by the cone $z^2=x^2+y^2$ and the sphere $x^2+y^2+z^2=a^2$ (this is the solid outside of the cone and inside the sphere.)
I got the triple integral $$2\int_0^{2\pi }\int_{\pi /4}^{3\pi /4} \int_0^{a} r^2dr\sin \theta d\theta d\phi .$$ Is this correct (I'm taking $\phi $ as the angle in the x-y plane)?

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Use Cylindrical Coordinates. The '$\color{#f00}{upper}$ volume' $\ds{V}$ is given by:

\begin{align}
V & \,\,\,\stackrel{\mrm{def.}}{=}\,\,\,
\iiint_{\mathbb{R}^{3}}\bracks{r^{2} + z^{2} > a^{2}}
\bracks{z^{2} < r^{2}}\bracks{\color{#f00}{z > 0}}r\,\dd r\,\dd\theta\,\dd z
\\[5mm] & =
2\pi\int_{0}^{\infty}\int_{0}^{\infty}\bracks{{\root{2} \over 2}\,a < r < a}
\bracks{\root{a^{2} - r^{2}} < z < r}r\,\dd r\,\dd z
\\[5mm] & =
2\pi\int_{\root{2}a/2}^{a}\pars{r - \root{a^{2} - r^{2}}}r\,\dd r =
\bbx{\ds{{1\over 3}\,\pars{2 - \root{2}}\pi a^{3}}}
\end{align}


TITLE: Find the value of complex expression $\left(\frac{\sqrt{3}+i}{2}\right)^{69}$
QUESTION [1 upvotes]: Find the value of

$$\left(\dfrac{\sqrt{3}+i}{2}\right)^{69}.\DeclareMathOperator{\cis}{cis}$$

I tried to solve this complex expression by converting it into polar form.
I expressed it in polar form $r\cis(t)$ from rectangular form $x+iy$ where $\cis(t) = \cos(t) + i\sin(t)$.
But I am unable to solve further due to the exponent of 69!

REPLY [3 votes]: Converting to polar coordinates is probably the best way to go.
$$r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=1$$
$$\theta = \arctan(1/\sqrt{3})=\frac{\pi}{6}$$
Then $$(re^{\theta i })^{69}=e^{\frac{69\pi}{6}i}=e^{10\pi i+\frac{3\pi}{2}i}=e^{\frac{3\pi}{2}i}=-i$$


TITLE: How do we get a simplicial homology functor?
QUESTION [5 upvotes]: The $n$-th simplicial homology group $H_n(A)$ of an abstract simplicial complex $A$ depends on the choice of an orientation for $A$ (but for different orientations, the homology groups are isomorphic!).
Does that mean that in order to get a functor $$H_n \;\colon\; \mathrm{Smp} \to \mathrm{Ab}$$ I have to choose an orientation for every possible abstract simplicial complex? If so, this will require a version of the axiom of choice for classes. Is this reasonable or is there a nicer way to get such a functor?
Edit (after first few answers):
Many people have proposed to include the orientation of a simplex in its definition (oriented simplex). This is a valid approch of course.
But I should have mentioned that I would like to use, if possible, the (minimalistic) definition of an abstract simplex, which you can find e.g. on Wikipedia. Sorry for not making this clear.

REPLY [2 votes]: The trick is to embed the category of abstract simplicial complexes inside the category of symmetric simplicial sets (= functor $\mathbf{F}^\mathrm{op} \to \mathbf{Set}$, where $\mathbf{F}$ is the category of positive finite cardinals): this can be done by sending an abstract simplicial complex $X$ to the symmetric simplicial set $\mathrm{Hom}(\Delta^{\bullet}, X)$.
Now let $\mathbf{\Delta}$ be the category of positive finite ordinals (and monotone maps). There is an evident embedding $\mathbf{\Delta} \to \mathbf{F}$, so by restriction, every symmetric simplicial set is also a simplicial set (= functor $\mathbf{\Delta}^\mathrm{op} \to \mathbf{Set}$). It is straightforward to define a homology functor $H_* : [\mathbf{\Delta}^\mathrm{op}, \mathbf{Set}] \to \mathbf{Gr Ab}$.
Of course, what one has to show is that putting all this together recovers the traditional definition of simplicial homology for an abstract simplicial complex. Let $\Delta^n$ be the standard $n$-simplex (with its canonical ordering), let $X$ be an abstract simplicial complex, let $Y$ be the corresponding symmetric simplicial set, and choose a linear ordering of the vertices of $X$. Then the chain complexes $C_{\bullet} (X)$ and $C_{\bullet} (Y)$ are defined as follows:

$C_{n} (X)$ is freely generated by the set of (non-degenerate) $n$-simplices of $X$, and the boundary map $C_{n} (X) \to C_{n-1} (X)$ is defined by the usual alternating sum.
$C_{n} (Y)$ is freely generated by the set of (possibly degenerate) simplicial maps $\Delta^n \to X$, and the boundary map $C_{n} (Y) \to C_{n-1} (Y)$ is defined by the alternating sum where the numbering is induced by the canonical ordering of $\Delta^n$ (not the linear order of the vertices of $X$).

The linear ordering of the vertices of $X$ induces a chain map $C_{\bullet} (X) \to C_{\bullet} (Y)$ that sends each (non-degenerate) $n$-simplex of $X$ to the unique order-preserving simplicial map $\Delta^n \to X$ whose image is that $n$-simplex. I claim that the induced homomorphism $H_* (X) \to H_* (Y)$ is an isomorphism. I do not personally know an elementary proof of this, but in principle it should be possible to prove directly that $C_{\bullet} (X) \to C_{\bullet} (Y)$ is a chain homotopy equivalence.


TITLE: Internal forces in an isolated system
QUESTION [2 upvotes]: I did a weird activity and now I am trying to figure out the physics behind it:
While sitting on a chair I kept my feet above the ground and tried to move the chair forward. I was able to. Initially, I thought that perhaps the center of mass won't be changing as I was pushed backwards. But I moved the chair for a while and saw that it had moved a considerable distance and I was still sitting on the chair. Hence the center of mass did change for sure. According to Newton's second law, there must be an external force acting on the system(me and the chair). But I am not able to figure out what force is it. I have a feeling that it is somewhere related to the fact that the system is not completely isolated but I am not sure how.
Note:
I am not in contact with the ground or any other object close to me. I have lifted my feet and have my hands on my lap.

REPLY [2 votes]: The movement you are performing is similar to hopping or jumping - see diagram below. It is much more difficult when you sit on the chair than when you stand, because you do not have as much flexibility, but it is essentially the same kind of movement. You could do even better by standing on the chair and tying it to your feet, then jumping on the chair as you would on the ground.

Using your arms (instead of your legs) as springs, you squat in the chair (b) then suddenly push yourself up and forward, dragging the chair with you when your arms are extended. It is slightly more effective if you hold yourself upright first (a) then swoop down before pushing all in the same movement. Swinging your legs also helps - when jumping you swing your arms and push with your legs; in the chair you swing your legs and push with your arms.
You could think of the chair as shoes - you push the ground through them, and because they are tied to your feet they leave the ground with you when you jump.
There are external forces acting here - you and the chair are not an isolated system. Contact forces with the ground - friction and normal reaction - as well as inertia/momentum, are essential - you cannot perform this kind of movement in space or when floating in water. You have to push down and back on the ground in order to get the ground to push upward and forward on you.


TITLE: Different definitions of boundary
QUESTION [1 upvotes]: Wikipedia gives three apparently equivalent definitions of a boundary:

the closure of S without the interior of $S: ∂S = \bar S$ \ $S^o$.
the intersection of the closure of S with the closure of its complement: $∂S = \bar S ∩ \overline {(X \ S)}$.
the set of points p of X such that every neighborhood of p contains at least one point of S and at least one point not of S.

The first two are equivalent because of something like this.
But how do I show the first two follow from the third? I don't even a rigorous symbolic definition to use in such a proof.

REPLY [2 votes]: Suppose $p$ is such that every neighborhood of $p$ contains at least one point of $S$, and at least one point not of $S$.
Then, by the first property (every neighborhood of $p$ contains at least one point of $S$), $p$ is a limit point of $S$, so $p \in \overline S$.
The second property shows that $p$ cannot be in the interior of $S$. Thus the third definition implies the first. (It also directly implies $p$ is in the closure of the complement of $S$, if you are shooting for demonstrating the second definition holds: "$p$ is a limit point of complement of $S$" $\Leftrightarrow$ "every neighborhood of $p$ contains a point not in $S$" $\Leftrightarrow$ "there does not exist a neighborhood of $p$ contained in $S$" ).
Basically these same words in a different order will show that the first definition implies the third. (and getting to and from the second/third definitions proceeds in a similar manner).


TITLE: Is there a mirror pair shape that can contain a light ray bundle indefinitely?
QUESTION [0 upvotes]: Suppose you have a bundle of light rays with origin around a point $x_0 \pm \epsilon_x$ and direction of about $v_0 \pm \epsilon_v$ .Is there a pair of mirrors separated by a gap like in the image such that the bundle could be trapped bouncing back and forth between the mirrors indefinitely?

REPLY [0 votes]: A two sheet hyperboloid of revolution reflects any light ray directed towards one focus, so that the reflection heads towards the other focus.  Note that the reflection then dies not go beyond the center of the hyperboloid.  If you start out between the sheets and headed towards one focus, the reflections bounce back and forth between the sheets and never gets past the center nor cones back out.  Draw it out in two dimensions (which a recall the light beam actually explores) and see what happens.


TITLE: Simplifying $\frac { \sqrt2 + \sqrt 6}{\sqrt2 + \sqrt3}$?
QUESTION [1 upvotes]: Is there anything else you can do to reduce it to something "nicer" other than multiplying it by $\dfrac {\sqrt3 - \sqrt2}{\sqrt3 - \sqrt2}$ and get $\sqrt 6 -2 + \sqrt {18} - \sqrt {12}$? The reason I think there's a nicer form is because the previous problem in the book was to simplify $\sqrt{3+ 2 \sqrt 2} - \sqrt{3 - 2 \sqrt 2}$, which simplifies nicely to $\sqrt{( \sqrt 2+1)^2} - \sqrt{  ( \sqrt 2 - 1)^2} = 2.$

REPLY [1 votes]: Well, you could note that
$$\sqrt6-2+\sqrt{18}-\sqrt{12}=-2+\sqrt6(1-\sqrt2+\sqrt3)$$
But beyond that, it doesn't look any better.


TITLE: Applying the Euler-Lagrange equations to Maxwell's Theory
QUESTION [3 upvotes]: In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
\mathcal{L} = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2
$$
and then he computes the following
$$
\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\partial_\mu A_\nu + (\partial_\rho A^\rho)\eta^{\mu\nu}.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!

REPLY [3 votes]: We have $\frac12 (\partial_{\mu}A^{\mu})^2 = \frac12 (\partial_{\alpha} A^{\alpha})(\partial_{\beta}A^{\beta})= \frac12 (\partial_{\alpha} A_{\sigma}) \eta^{\sigma\alpha}(\partial_{\beta}A_{\rho}) \eta^{\rho\beta}$ so the derivative w.r.t. $\partial_{\mu} A_{\nu}$ is
$$\frac12\delta_{\alpha}^{\mu} \delta_{\sigma}^{\nu} \eta^{\sigma\alpha}(\partial_{\beta}A_{\rho}) \eta^{\rho\beta}+\frac12(\partial_{\alpha} A_{\sigma}) \eta^{\sigma\alpha}\delta_{\beta}^{\mu} \delta_{\rho}^{\nu} \eta^{\rho\beta}= \frac12 \eta^{\mu\nu} (\partial_{\beta} A^{\beta})+\frac12 (\partial_{\alpha}A^{\alpha}) \eta^{\mu\nu} = (\partial_{\rho}A^{\rho})\eta^{\mu\nu} $$
where I've freely labeled and relabeled dummy indices.


TITLE: Temperature and kinetic energy of molecules
QUESTION [1 upvotes]: I was wondering if temperature is related to the average translational kinetic energy of the molecules then why does the average kinetic energy of say a moving object not affect the temperature? What is the mechanism regarding these vibrational motions that imparts temperature. Also do a gas and liquid at the same temperature possess the same kinetic energy? and a heavier molecule traveling at the same speed as a lighter molecule will have a greater temperature?

REPLY [1 votes]: This link will help  s will the image :
.
Temperature is a thermodynamic variable, i.e. at the atomic and molecular level it has to do with the kinetic energy of the atoms/molecules/lattices , whose average can be shown to be connected to  the thermodynamic temperature.
For an ideal gas :

why does the average kinetic energy of say a moving object not affect the temperature

Because  it is at the rest frame of the objects that temperature is defined. The overall motion makes no difference. ( Are you hotter when moving with 60 miles an hour?)

What is the mechanism regarding these vibrational motions that imparts temperature.

Molecules can contain extra motions , as seen in the link, and this can add extra complication in the statistical estimate.  Usually temperature goes by the thermodynamic definition.
As quantum mechanics is the framework for correct estimates at the microscopic level, the QM energy levels of vibration and rotation have to be taken into account,as well as the scatterings and radiation due to the inherent charges in  the material. This leads to black body radiation which is specific to the temperature of the sample, but that is another story.
As temperature can be equated with an average kinetic energy, if the kinetic energy of a molecule is measured, one could say that it corresponds to the average kinetic energy of temperature T, but it has no meaning to say that a molecule has temperature, as to get thermodynamics one needs a large  number of molecules.


TITLE: A question about Elementary Row Operation: Add a Multiple of a Row to Another Row
QUESTION [1 upvotes]: The task is that I have to prove the following statement, using Linear Algebra arguments:

Given a matrix A, then:
To perform an ERO (Elementary Row Operation) type 3 :
(c * R_i)  + R_k --> R_k  (i.e.  replace a row k by adding c-multiple of row i to row k)
is the same as replacing a row k by subtracting a multiple of some row from another row

I just don't know how to formally prove this statement, like how the arguments should look like.
By some inspections, I'm pretty sure that doing

(c * R_i) + R_k --> R_k

is the same as doing:

R_i - (d * R_k) --> R_k

where d can be positive or negative, but it must have opposite sign with c.  I use an example as follows:
A = (2 1 3, 4 3 1)
Then if I want to add row 2 to row 1, say, instead of doing (1 * 4) + 2 --> 6 and so on, I do 4 - [ (-1) * 2 ] --> 6 instead.  Thus c = 1 and d = -1 in this case.
That's why I conclude that the coefficient d should always be the opposite sign with coefficient c.
Would someone help me on how to construct a formal proof of the statement?  I know how to go about the examples, but I understand examples are never proofs >_<
Thank you very much ^_^

REPLY [1 votes]: Note: If $c$ is some non-zero scalar, then

adding $cR_i$ to $R_k$ and replacing the original $R_{k\text{ old}}$ by $(R_k + cR_i)$

is the same as

subtracting $−c⋅R_i$ from $R_k$ and replacing the old $R_k$ by the result $R_k - (-cR_i)$.

Since...$R_k + cR_i = R_k - (-cR_i)$


TITLE: Propositional Logic Exercise
QUESTION [0 upvotes]: I need your help in solving an exercise in propositional logic:
I am trying to prove that this :
(Y ∧ ¬Z) ∨ ((X ∨ Y) ∧ (¬Y ∨ X))

is equivalent to this (according to the solution of the exercise)
(X ∨ ¬Z) ∧ (X ∨ Y)

What I did so far:
(Y ∧ ¬Z) ∨ ((X ∨ Y) ∧ (¬Y ∨ X)) ≡ (Y ∧ ¬Z) ∨ (( X ∧ (¬Y ∨ X)) ∨ ( Y ∧ (¬Y ∨ X))
 ≡ (Y ∧ ¬Z) ∨ (( X ∧ ¬Y) ∨ (X ∧ X) ∨ (Y ∧ ¬Y) ∨ (Y ∧ X))
 ≡ (Y ∧ ¬Z) ∨ ( X ∧ ¬Y) ∨ X ∨ (Y ∧ X)
 ≡ (Y ∧ ¬Z) ∨ X ∨ ( X ∧ ¬Y) ∨ (Y ∧ X)
 ≡ ((X ∨ Y) ∧ (X ∨ ¬Z)) ∨ ( X ∧ ¬Y) ∨ (Y ∧ X)

Now this part what I am having issues with, I could apply the distributive law again, but I don't see how this part will turn False at some point. :
( X ∧ ¬Y) ∨ (Y ∧ X) ≡ ( X ∨ (Y ∧ X)) ∧ ( ¬Y ∨ (Y ∧ X))
                    ≡ (X ∨ Y) ∧ (X ∨ X) ∧ (¬Y ∨ Y) ∧ (¬Y ∨ X)
                    ≡ (X ∨ Y) ∧ X ∧ Y ∧ (¬Y ∨ X)

Thank you for your time and help!

REPLY [1 votes]: We have to apply Distributivity to $(X ∨ Y) ∧ (¬Y ∨ X)$ to get : $X ∨ (Y ∧ ¬Y)$.
And $(Y ∧ ¬Y) \equiv \text F$ and $X ∨ \text F \equiv X$.
Thus, starting from the original formula :

$$[(Y ∧ ¬Z) ∨ ((X ∨ Y) ∧ (¬Y ∨ X))] \equiv [(Y ∧ ¬Z) ∨ (X ∨ (Y ∧ ¬Y))] \equiv$$
$$[(Y ∧ ¬Z) ∨ (X ∨ \text F)] \equiv [(Y ∧ ¬Z) ∨ X].$$


TITLE: Why is 1 the derivative of $x$?
QUESTION [1 upvotes]: Can someone explain to me why the derivative of $x$ is 1. We just started derivatives in my calculus class and I don't understand it at all.

REPLY [0 votes]: You can use the power rule for finding the derivative that,$$\frac{dx^n}{dx}= n x^{n-1}.$$
We know that $x$ is the same thing as $x^1$, and use the above power rule, you get $\frac{dx}{dx}=1 x^0$ and we also know that $x^0$ is $1$, so $\frac{dx}{dx}=1 * 1=1$.
The integral is $$\int x^n dx= \frac{x^{n+1}}{n+1}$$
So the derivative of $x$ is one and the integral of $dx$ is $x$.  If you want to go little different, you can use this formula for finding the derivative, $$\frac{d f(x)}{dx}=
\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}h$$ you can substitute $x$ in $f(x)$ and still you will get one.


text:
\begin{document}

\title*{Deformation quantization on a Hilbert space}

\author{Giuseppe Dito}

\institute{Institut de Math\'ematiques de Bourgogne\\ Universit\'e de Bourgogne\\
UMR CNRS 5584\\ B.P. 40780,
21078 Dijon Cedex, France\\
\texttt{giuseppe.dito@u-bourgogne.fr} }

\maketitle

\abstract{
We study deformation quantization on an infinite-dimensional
Hilbert space $W$ endowed with its canonical Poisson structure.
The  standard example of the Moyal star-product is made explicit
and it is shown that it is well defined on a subalgebra of $C^\infty(W)$.
A classification of inequivalent deformation quantizations of exponential type,
containing the Moyal and normal star-products, is also given.}

\section{Introduction}\label{sec:intro}

Deformation quantization provides an alternative formulation of Quantum
Mechanics by interpreting quantization as a deformation of the
commutative algebra a classical observables into a noncommutative
algebra \cite{bayen.et.al:1978a}. The quantum algebra is defined by
a formal associative star-product $\star_\hbar$ which
encodes the algebraic structure of the set of observables.

Deformation quantization has been applied with increasing generality
to several area of mathematics and physics. Most of these applications
deal with star-products on finite-dimensional manifolds.
See \cite{dito.sternheimer:2002a} for a recent review.

It is natural to consider an extension of deformation quantization to
infinite-dimensional manifolds as it appears to be a good setting where
quantum field theory of nonlinear wave equations
can be formulated (e.g. in the sense of I.~Segal \cite{segal:1974a}).
In the star-product approach, the first steps in that direction are given in \cite{dito:1990a,dito:1992a}.

Recently, deformation quantization has become popular among  field and string theorists.
A generalization of Moyal star-product to infinite-dimensional spaces
appears in several places in the literature.
Let us just notice that the  Witten star-product \cite{witten:1986} appearing
in string field theory is {\em heuristically} equivalent to
an infinite-dimensional version of the Moyal star-product.
A brute force generalization of Moyal star-product to field theory
yields to some pathological and unpleasant features
such as anomalies and breakdown of associativity.
We think that it is worth writing down a mathematical study of the Moyal
product in infinite dimension even if it is not an adequate product
for field theory considerations.

In the finite-dimensional case, the existence of star-products on any (real) symplectic manifold
has been established by DeWilde and Lecomte \cite{dewilde.lecomte:1983b}. The general existence and classification
problems for the deformation quantization of a Poisson manifold was solved by Kontsevich \cite{kontsevich:1997}.
However, the very first problem that one faces when going over infinite-dimensional spaces,
it to make sense of the star-product itself as a formal associative product.
It contrasts with the finite-dimensional case where the deformation is defined
on all of the smooth functions on the manifold. This is by far too demanding
in the infinite-dimensional case even when the Poisson structure is well-defined
on all of the smooth functions (e.g. on Banach or Fr\'echet spaces).
One should specify first an Abelian algebra of admissible functions which then
can be deformed. For example, on $E=\mathcal{S}\times\mathcal{S}$, where $\mathcal{S}$ is the
Schwartz space on $\R^n$, endowed with its canonical Poisson structure,
one cannot expect to write down a star-product defined on all holomorphic
functions on $E$, but has to restrict it to some subalgebra. For example, in \cite{dito:1992a} it is shown,
that for such a simple star-product as the normal star-product, it is defined on the subalgebra
of holomorphic functions of $a$ and $\bar a$ (creation and annihilation `operators') having
semi-regular kernels. In \cite{duetsch.fredenhagen:2001b}, one can find a nice  analysis for the normal
star-product and the conditions on the kernel have been translated in terms of  wave front set of the distributions.

After making precise what is a deformation quantization on a Hilbert space,
we first present a study of Moyal product when the space-space
is the direct sum of a Hilbert space with its dual.
We identify a subalgebra of smooth functions, specified by conditions of Hilbert-Schmidt
type on their derivatives, on which the Moyal product makes sense.
We also define a family of star-products of exponential type, show that
they are not all equivalent to each other and give the classification
of their equivalence classes in terms of Hilbert-Schmidt operators.

\section{Star-products on a Hilbert space}

When  infinite-dimensional spaces are involved, further conditions
are needed to define a deformation quantization or a star-product.
The algebra of functions on which the Poisson bracket and the star-product are
defined should be specified along with the class of {\em admissible} cochains
(especially when the issue of the equivalence of deformations is considered).

\subsection{Notations}

Let $B$ be a Banach space over a field $\K$ ($\R$ or $\C$). The topological dual of $B$
shall be denoted by $B^*$.
The Banach space  of bounded $r$-linear  forms on~$B$ is denoted by $\mathcal{L}^r(B,\K)$
and $\mathcal{L}^r_{\rm sym}(B,\K)$ is the subspace of $\mathcal{L}^r(B,\K)$ consisting
of bounded symmetric $r$-linear forms on $B$.
We shall denote by $C^\infty(B,\K)$ the space of $\K$-valued functions on $B$ that
are smooth in the Fr\'echet sense.  The Fr\'echet derivative of $F\in C^\infty(W,\K)$ is denoted by $DF$ and
it is a smooth map from $B$ to $\mathcal{L}^1(B,\K)=B^*$, i.e., $DF\in C^\infty(B,B^*)$.
For $F\in C^\infty(B,\K)$, the higher derivative $D^{(r)}F$ belongs to $C^\infty(B,\mathcal{L}^r_{\rm sym}(B,\K))$
and we shall use the following notation $D^{(r)}F(b).(b_1,\ldots,b_r)$ for the $r^{\rm th}$-derivative
of $F$ evaluated at $b\in B$ in the direction of $(b_1,\ldots,b_r)\in B^r$.

Let $W$ be an infinite-dimensional separable Hilbert space over a field  $\K$.
For notational reasons, as it will become clear later, it would be convenient
for us to not identify $W^*$ with $W$.
For any orthonormal basis $\{e_i\}_{i\geq1}$ in $W$ and corresponding
dual basis $\{e^*_i\}_{i\geq1}$ in $W^*$,
we shall denote the partial derivative of $F\in C^\infty(W,\K)$ evaluated at $w$
in the direction of $e_i$ by $\partial_iF(w)\in \K$, i.e., $\partial_iF(w)=DF(w).e_i$.
Since $F$ is differentiable in the  Fr\'echet sense, we have $DF(w) = \sum_{i\geq1} \partial_iF(w)e^*_i$
and thus, for any $w\in W$, that $\sum_{i\geq1} |\partial_iF(w)|^2<\infty$.

\subsection{Multidifferential operators}

Let us first make precise what we call a Poisson
structure on $W$. In the following, we will consider  a map $P$ that sends $W$ into a space
of (not necessarily bounded) bilinear forms on $W^*$ and a subalgebra $\F$ of $C^\infty(W,\K)$.
We define the subspace $\mathcal{D}^\F_w =\{DF(w)\ | \ F\in \F \}$ of  $W^*$.
\vskip2mm
\begin{definition}\label{def:poisson}
Let $W$ be a Hilbert space. Let $\F$ be an Abelian subalgebra (for the pointwise product)
of $C^\infty(W,\K)$.
A Poisson bracket on $(W,\F)$ is a $\K$-bilinear map $\{\cdot,\cdot\}\colon \F\times \F \rightarrow \F$
such that:
\vskip2mm
\noindent i) there exists a map $P$ from $W$ to the  space of bilinear forms on $W^*$,
so that the domain of $P(w)$ contains $\mathcal{D}^\F_w\times \mathcal{D}^\F_w$ and
$\forall F,G\in \F$, $\{F,G\}(w) = P(w).(DF(w),DG(w))$ where $w\in W$.
\vskip2mm
\noindent ii) $(\F, \{\cdot,\cdot\})$ is a Poisson algebra,
i.e., skew-symmetry, Leibniz rule, and Jacobi identity are satisfied.
\vskip2mm
\noindent The triple $(W,\F,\{\cdot,\cdot\})$ is called a Poisson space.
\end{definition}
\vskip2mm
Let us give an example where $P(w)$ is an unbounded bilinear form on $W^*$.

\begin{example}
Consider a real Hilbert space $W$ with orthonormal basis $\{e_i\}_{i\geq0}$.
We will realize the following subalgebra of the Witt algebra:
$$
[L_m,L_n] = (m-n) L_{m+n},\quad m,n\geq 0,
$$
by functions on $W$. For $w\in W$, let $\phi_i(w)=\langle e^*_i, w\rangle$, $i\geq0$ be the coordinate
functions. The algebra $\mathcal F$ generated by the family of functions $\{\phi_i\}_{i\geq0}$
is an Abelian  subalgebra of $C^\infty(W,\R)$ consisting of polynomial functions in a finite number
of variables. The following expression:
$$
\{F,G\}(w) = \sum_{m,n\geq0} (m-n) \phi_{m+n}(w) \partial_m F(w) \partial_n G(w),\quad F,G\in \F, w \in W,
$$
defines a Poisson bracket on $(W,\F)$. Indeed the right-hand side is a finite sum and is a function
in  $\F$, and we have:
$$
\{\phi_i,\phi_j\}=(i-j)\phi_{i+j},
$$
from which Jacobi identity follows. The special case when $j=0$ gives:
$$
\{\phi_i,\phi_0\}(w)=P(w).(D\phi_i(w), D\phi_0(w)) = P(w).(e^*_i,e^*_0)=  i \phi_{i}(w).
$$
By choosing an appropriate $w$ (e.g. $w=\sum_{i\geq1} i^{-3/4} e_i$), then
$P(w).(e^*_i,e^*_0)=i \phi_{i}(w)$ can become as large
as desired by varying $i$. This shows that the bilinear form $P(w)$ cannot be bounded.
\end{example}

The generalization of Def.~\ref{def:poisson} to multidifferential operators on $W$ is straightforward.
Again, given an Abelian subalgebra $\F$ of $C^\infty(W,\K)$, we define the following subspace of
$\mathcal{L}^r_{\rm sym}(W,\K)$:
$$
\mathcal{D}^\F_w(r) =\{D^{(r)}F(w)\ | \ F\in \F \}.
$$
\vskip2mm
\begin{definition}\label{def:op}
Let $W$ be a Hilbert space. Let $\F$ be an Abelian subalgebra, for the pointwise product,
 of $C^\infty(W,\K)$.
Let $r\geq1$, an $r$-differential operator ${\bf A}$ on $(W,\F)$ is an
$r$-linear map ${\bf A} \colon \F^r \rightarrow \F$
such that:
\vskip2mm
\noindent i) for $(n_1,\ldots,n_r)\in\N^r$, there exists a  map $a^{(n_1,\ldots,n_r)}$ from
$W$ to a space of (not necessarily bounded) $r$-linear forms on
$\mathcal{L}^{n_1}_{\rm sym}(W,\K)\times\cdots\times \mathcal{L}^{n_r}_{\rm sym}(W,\K)$, i.e.,
$$
a^{(n_1,\ldots,n_r)}(w)\colon \mathcal{D}_w^{(n_1,\ldots,n_r)}\subset
\mathcal{L}^{n_1}_{\rm sym}(W,\K)\times\cdots\times \mathcal{L}^{n_r}_{\rm sym}(W,\K)\rightarrow \K,
$$
so that the domain $\mathcal{D}_w^{(n_1,\ldots,n_r)}$ of $a^{(n_1,\ldots,n_r)}(w)$ contains
$\mathcal{D}^\F_w(n_1)\times\cdots \times\mathcal{D}^\F_w(n_r)$
and $a^{(n_1,\ldots,n_r)}$ is $0$ except for finitely many $(n_1,\ldots,n_r)$;
\vskip2mm
\noindent ii) for any $F_1,\ldots,F_r\in \F$ and $w\in W$, we have
$$
{\bf A}(F_1,\ldots,F_r)(w)=\sum_{n_1,\ldots,n_r\geq 0} a^{(n_1,\cdots ,n_r)}(w).(D^{(n_1)}F_1(w),\cdots,D^{(n_r)}F_r(w)).
$$
\end{definition}
\vskip2mm
Notice that Poisson brackets as defined above are special cases of
bidifferential operators in the sense of Def.~\ref{def:op} with $P=a^{(1,1)}$.

\subsection{Deformation quantization on $W$}

We now have all the ingredients to define what is meant by deformation quantization
of a Poisson space $(W,\F,\{\cdot,\cdot\})$ when $W$ is a Hilbert space.
\vskip2mm
\begin{definition}
Let $W$ be a Hilbert space and $(W,\F,\{\cdot,\cdot\})$ be a Poisson space.
A star-product on $(W,\F,\{\cdot,\cdot\})$ is a $\K[[\hbar]]$-bilinear product
$\star_\hbar\colon \F[[\hbar]]\times\F[[\hbar]]\rightarrow\F[[\hbar]]$ given by
$F\star_\hbar G = \sum_{r\geq0} \hbar^r C_r(F,G)$ for $F,G\in\F$ and extended by $\K[[\hbar]]$-bilinearity
to $\F[[\hbar]]$, and satisfying for any
$F,G,H\in\F$:
\vskip2mm
\noindent i)
$C_0(F,G)= FG$,
\vskip2mm
\noindent ii)
$C_1(F,G) - C_1(G,F) = 2 \{F,G\}$,
\vskip2mm
\noindent iii)
for $r\geq1$, $C_r\colon\F\times\F\rightarrow \F$ are bidifferential operators in the sense of Def.~\ref{def:op},
vanishing on constants,
\vskip2mm
\noindent iv)
$F\star_\hbar(G\star_\hbar H)=(F\star_\hbar G)\star_\hbar H$.
\vskip2mm
\noindent The triple $(W,\F[[\hbar]],\star_\hbar)$ is called a deformation quantization
of the Poisson space $(W,\F,\{\cdot,\cdot\})$.
\end{definition}
\vskip2mm
We also have a notion of equivalence of deformations adapted to our context:
\vskip2mm
\begin{definition}
Two deformation quantizations $(W,\F[[\hbar]],\star^1_\hbar)$ and $(W,\F[[\hbar]],\star^2_\hbar)$
of the same Poisson space $(W,\F,\{\cdot,\cdot\})$ are said to be equivalent if
there exists a $\K[[\hbar]]$-linear map $T\colon\F[[\hbar]]\rightarrow\F[[\hbar]]$ expressed
as a formal series $T=\mathrm{Id}_\mathcal{F} + \sum_{r\geq1} \hbar^r T_r$ satisfying:
\vskip2mm
\noindent i) $T_r\colon \F\rightarrow\F $, $r\geq1$, are differential operators in the sense of Def.~\ref{def:op},
vanishing on constants,
\vskip2mm
\noindent ii) $T(F)\star^1_\hbar T(G) = T(F \star^2_\hbar G),\ \forall F,G\in \F$.
\end{definition}

\section{Moyal product on a Hilbert space}\label{sec:toy}

We present an infinite-dimensional version of the Moyal product defined on a class
of smooth functions specified by a Hilbert-Schmidt type of conditions on their derivatives.

\subsection{Poisson structure}\label{sub:poisson}

Let $\H$ be an infinite-dimensional separable Hilbert space.
We consider the phase-space $W=\H\oplus\H^*$ endowed with its canonical strong symplectic
structure $\omega((x_1,\eta_1),(x_2,\eta_2))= \eta_1(x_2) - \eta_2(x_1)$, where
$x_1,x_2\in\H$ and $\eta_1,\eta_2\in\H^*$.

Let $F\colon W\rightarrow \mathbb{C}$ be a $C^\infty$ function (in the Fr\'echet sense).
We shall denote by $D_1F(x,\eta)$ (resp. $D_2F(x,\eta)$) the first (resp. second) partial Fr\'echet derivative of $F$
evaluated at point $(x,\eta)\in W$. With the identification
$\H^{**}\sim\H$ we have  $D_1F(x,\eta)\in \H^*$
and $D_2F(x,\eta)\in \H$. Let $\langle\cdot, \cdot \rangle\colon\H^* \times\H\rightarrow\K$ be the canonical
pairing between $\H$ and $\H^*$.

With these notations, the bracket associated with the
canonical symplectic structure on $W$ takes the form:
\begin{equation}\label{poisson}
\{F,G\}(x,\eta)
=\langle D_1F(x,\eta),D_2G(x,\eta)\rangle-\langle  D_1G(x,\eta),D_2F(x,\eta)\rangle,
\end{equation}
where $F,G\in C^\infty(W,\K)$.
\vskip2mm
\begin{proposition}\label{prop:poisson}
The space $W$ endowed with the bracket (\ref{poisson}) is an infinite-dimen\-sional
Poisson space or, equivalently, $(C^\infty(W,\K), \{\cdot,\cdot\})$ is a Poisson algebra.
\end{proposition}

\begin{proof} One has only to check that the map $(x,\eta)\mapsto \{F,G\}(x,\eta)$ belongs
to $C^\infty(W,\K)$ for any $F,G\in C^\infty(W,\K)$. Then Leibniz property and Jacobi identity will follow.
For $F,G\in C^\infty(W,\K)$, the maps $(x,\eta)\mapsto (D_1F(x,\eta),D_2G(x,\eta))$ and
$(\xi, y) \mapsto \langle \xi, y \rangle$ belong to $C^\infty(W,\H^*\times\H)$
and $C^\infty(\H^*\times\H,\K)$,  respectively. The map $(x,\eta)\mapsto \{F,G\}(x,\eta)$, as composition of
$C^\infty$ maps, is therefore in $C^\infty(W,\K)$.\hfill$\square$
\end{proof}
\vskip2mm
For any orthonormal basis $\{e_i\}_{i\geq1}$ in $\H$ and dual basis $\{e^*_i\}_{i\geq1}$ in $\H^*$,
the complex number $\partial_iF(x,\eta)$ shall denote the partial derivative of $F$ evaluated at $(x,\eta)$
in the direction
of $e_i$, i.e. $\partial_iF(x,\eta)=DF(x,\eta).(e_i,0)=D_1F(x,\eta).e_i$, and, similarly,
$\partial_{i^*}F(x,\eta)=DF(x,\eta).(0,e_i^*)=D_2F(x,\eta).e_i^*$
is the partial derivative in the direction of $e_i^*$. Notice that $i^*$ should not be considered
as a different index from~$i$ when sums are involved, it is merely a mnemonic notation to distinguish
partial derivatives in $\H$ and in $\H^*$.

For $F\in C^\infty(W,\K)$,
we have for any $(x,\eta)\in W$ that $\sum_{i\geq1} |\partial_iF(x,\eta)|^2<\infty$ and
$\sum_{i\geq1} |\partial_{i^*}F(x,\eta)|^2<\infty$, hence
the Poisson bracket~(\ref{poisson}) admits an equivalent form in terms of
an absolutely convergent series:
\begin{equation}\label{poissonbis}
\{F,G\}(x,\eta)  = \sum_{i\geq1}
\big(\partial_iF(x,\eta) \partial_{{i}^*}G(x,\eta)-\partial_iG(x,\eta) \partial_{{i}^*}F(x,\eta)\big).
\end{equation}

\subsection{Functions of Hilbert-Schmidt type}\label{sub:hs}

We now define a subalgebra of $C^\infty(W,\K)$ suited for our discussion.
Let us start with some definitions and notations.

For any $F\in C^\infty(W,\K)$ and  $(x,\eta)\in W$, the higher derivatives
$$
D^{(r)}F(x,\eta)\colon W\times\cdots\times W\rightarrow \K,\quad r\geq 1,
$$
are bounded symmetric $r$-linear maps and
partial derivatives of $F$ will be denoted
$D^{(r)}_{\alpha_1 \cdots \alpha_r}F(x,\eta)$ where $\alpha_1,\ldots,\alpha_r$ are taking values $1$ or $2$.
Let us introduce:
\begin{equation}\label{defH}
\H^{(\alpha)} = \begin{cases}
\H,  & \mbox{if}\ \alpha=1;\\
\H^*  ,  & \mbox{if}\ \alpha=2.
\end{cases}
\qquad
\alpha^\flat = \begin{cases}
2 , & \mbox{if}\ \alpha=1;\\
1 ,   & \mbox{if}\ \alpha=2.
\end{cases}
\qquad
i^{(\alpha)} = \begin{cases}
i , & \mbox{if}\ \alpha=1;\\
i^* ,   & \mbox{if}\ \alpha=2.
\end{cases}
\end{equation}
Also $i^\sharp$ will stand for either $i$ or $i^*$.
With these notations, partial derivatives of $F$ are bounded $r$-linear maps:
$$
D^{(r)}_{\alpha_1 \cdots \alpha_r}F(x,\eta)\colon \H^{(\alpha_1)}\times\cdots\times \H^{(\alpha_r)}\rightarrow \K.
$$

It is convenient to introduce new symbols such as $\partial_{ij^*k}$
for higher partial derivatives, e.g., $\partial_{ij^*k}F(x,\eta)\in \K$ stands for
$D^{(3)}F(x,\eta).((e_i,0),(0,e_j^*),(e_k,0))=D^{(3)}_{121}F(x,\eta).(e_i,e_j^*,e_k)$,
where $\{e_i\}_{i\geq1}$ (resp. $\{e_i^*\}_{i\geq1}$)
is an orthonormal basis in $\H$ (resp. $\H^*$).
\vskip2mm
\begin{definition}\label{defhs} Let $\{e_i\}_{i\geq1}$
be an orthonormal basis in $\H$ and $\{e_i^*\}_{i\geq1}$ be the dual basis in $\H^*$.
Functions of Hilbert-Schmidt type are functions $F$ in $C^\infty(W,\K)$ such that

\begin{equation}\label{hs}
\sum_{i_1, \ldots, i_r \geq1} |\partial_{i^{\sharp}_1\cdots i^{\sharp}_r}F(x,\eta)|^2 < \infty,
\quad \forall r\geq1, \forall (x,\eta)\in W.
\end{equation}

The sums involved have to be  interpreted in the sense of  summable families.
By Schwarz lemma for partial derivatives, it should be understood that Eq.~(\ref{hs})
represents $r+1$ distinct sums corresponding to all of the choices $i^\sharp=i$ or $i^*$.
The set of functions of Hilbert-Schmidt type on $W$ will be denoted by $\F_{HS}$.
\end{definition}
\vskip2mm
The definition above is independent of the choice of the orthonormal basis.
\vskip2mm
\begin{remark}
Let $\N_*$ be the set of positive integers. For each $r\geq1$,
the set of families of elements $\{x_I\}_{I\in \N_*^r}$ in $\K$
such that $\sum_{I\in \N_*^r}{|x_I|^2}<\infty$ is the
Hilbert space $\ell^2({\N_*^r)}$ for the usual operations and inner product.
Then condition~(\ref{hs}) can be equivalently stated in the following way:
$F\in C^\infty(W,\K)$ is of Hilbert-Schmidt type if and only if,
for any $r\geq1$ and any $(x,\eta)\in W$, the $2^r$ families
$\{\partial_{i^{\sharp}_1\cdots i^{\sharp}_r}F(x,\eta)\}_{(i^{\sharp}_1,\ldots, i^{\sharp}_r)\in  \N_*^r}$
belong to $\ell^2({\N_*^r)}$.
\end{remark}

\begin{remark}
The set $\F_{HS}$ does not contain all of the (continuous) polynomials on $W$. For
example, the polynomial $P(y,\xi)=\langle\xi, y\rangle$ is not in $\F_{HS}$
as $\sum_{i,j\geq1} |\partial_{ij^*}P(x,\eta)|^2$ $= \sum_{i,j\geq1} \delta_{ij}=\infty$.
In a quantum field theory context, the polynomial~$P$ corresponds to a free Hamiltonian
in the holomorphic representation.
\end{remark}
\vskip2mm
\begin{proposition}\label{pro:subhs}
The set of functions of Hilbert-Schmidt type is an Abelian subalgebra of $C^\infty(W,\K)$
for the pointwise product of functions.
\end{proposition}
\begin{proof}
Let $a,b\in \K$ and $F,G\in \F_{HS}$. It is clear from Remark~1 that
$aF + bG$ is in $F,G\in \F_{HS}$. The product $FG$ belongs to
$\F_{HS}$ as a consequence of the Leibniz rule for the  derivatives and from:
if $\{x_I\}_{I\in \N_*^r}\in \ell^2({\N_*^r)}$ and $\{y_J\}_{J\in \N_*^s}\in \ell^2({\N_*^s)}$,
then $\{x_Iy_J\}_{I\times J\in \N_*^{r+s}}\in \ell^2({\N_*^{r+s})}$. \hfill$\square$
\end{proof}
\vskip2mm
Moreover the Poisson bracket~(\ref{poisson}) restricts to $\F_{HS}$ and we have:
\vskip2mm
\begin{proposition}\label{pro:poihs}
$(W,\F_{HS}, \{\cdot,\cdot\})$ is a Poisson space.
\end{proposition}
\begin{proof}
Let $F$ and $G$ be in $\F_{HS}$. According to the proof of Prop.~\ref{prop:poisson}, the
map $\Phi\colon (x,\eta)\mapsto \langle D_1F(x,\eta),D_2G(x,\eta)\rangle$ is in $ C^\infty(W,\K)$
and splits as follows:

\begin{equation*}
\begin{CD}
\Phi\colon W @>{\Psi_1}>>\H^*\times \H  @>{\Psi_2}>>\K
\end{CD}
\end{equation*}

$$
(x,\eta)\mapsto (D_1F(x,\eta),D_2G(x,\eta)) \mapsto \langle D_1F(x,\eta),D_2G(x,\eta)\rangle,
$$
\vskip2mm
\noindent where both  $\Psi_1$ and  $\Psi_2$ are $C^\infty$ maps. We only need to check
that $\Phi$ is of Hilbert-Schmidt type.

By applying the chain rule
to $\Phi=\Psi_2\circ \Psi_1$, it is easy to see that we can freely interchange partial derivatives
with the sum sign and we get that the partial derivatives of $\Phi$ is a
finite sum of terms of the form:
\begin{equation}\label{summ}
a_{j_1^\sharp\cdots j_r^\sharp,k_1^\sharp\cdots k_s^\sharp}
\equiv\sum_{i\geq1} \partial_{ij_1^\sharp\cdots j_r^\sharp}F(x,\eta)
\partial_{i^*k_1^\sharp\cdots k_s^\sharp}G(x,\eta).
\end{equation}
The Cauchy-Schwarz inequality implies that the family
$\{ a_{j_1^\sharp\cdots j_r^\sharp,k_1^\sharp\cdots k_s^\sharp}\}$ is in $\ell^2(\N^{r+s}_*)$ and thus
$\Phi$ belongs to $\F_{HS}$. Hence $\F_{HS}$ is closed under the Poisson bracket.\hfill$\square$
\end{proof}

\subsection{Moyal star-product on $W$}\label{sub:moyal}

We are now in position to define the Moyal star-product on $W$ as an associative product
on $\F_{HS}[[\hbar]]$.

For $F,G\in \F_{HS}$, $(x,\eta)\in W$, $r\geq1$, $\alpha_1,\ldots,\alpha_r,\beta_1\ldots,\beta_r$ equal to
$1$ or $2$, and with the notations introduced previously, let us define:
\begin{equation}\label{conr}
\langle\langle D^{(r)}_{\alpha_1\cdots \alpha_r}F,D^{(r)}_{\beta_1\cdots \beta_r}G\rangle\rangle(x,\eta)
= \sum_{i_1,\ldots,i_r\geq1}
\partial_{i^{(\alpha_1)}_1\cdots i^{(\alpha_r)}_r}F(x,\eta)\
\partial_{i^{(\beta_1)}_1\cdots i^{(\beta_r)}_r}G(x,\eta).
\end{equation}
\vskip2mm
\begin{remark}
The preceding definition does not depend on the choice of the orthonormal basis in $\H$ and
the series is absolutely convergent as a consequence of the Cauchy-Schwarz inequality.
\end{remark}
\vskip2mm
Let $\Lambda$ be the canonical symplectic $2\times2$-matrix
with $\Lambda^{12}=+1$.
As in the finite-dimensional case, the powers of the Poisson bracket~(\ref{poisson}) are defined as:
\begin{equation}\label{eq:cr}
C_r(F,G)
 = \sum_{\alpha_1,\ldots,\alpha_r=1,2}\sum_{\beta_1,\ldots,\beta_r=1,2}
\Lambda^{\alpha_1\beta_1}\cdots \Lambda^{\alpha_r\beta_r} \langle \langle D^{(r)}_{\alpha_1\cdots \alpha_r}F,
D^{(r)}_{\beta_1\cdots \beta_r}G\rangle\rangle.
\end{equation}

The next Proposition shows that the $C_r$ are bidifferential operators in the sense of Def.~\ref{def:op}
and they close on $\F_{HS}$. We shall use a specific version of
the Hilbert tensor product $\otimes$ between Hilbert spaces
(see e.g. Sect.~2.6 in \cite{kadison.ringrose:1997a}). Let $\H_1,\ldots,\H_r$ be Hilbert
spaces with orthonormal bases $\{e^{(1)}_i\}_{i\geq1},\ldots,\{e^{(r)}_i\}_{i\geq1}$.
There exists a Hilbert space $\T=\H_1\otimes \cdots \otimes\H_r$ and
a bounded $r$-linear map $\Psi\colon(x_1,\ldots,x_r)\mapsto x_1\otimes \cdots \otimes x_r$
from  $\H_1\times \cdots \times\H_r$ to $\H_1\otimes \cdots \otimes\H_r$ satisfying:
$$
\sum_{i_1,\ldots,i_r\geq1} |\langle\Psi(e^{(1)}_{i_1},\ldots,e^{(r)}_{i_r}),x\rangle|^2 <\infty, \quad \forall x\in \T,
$$
such that for any bounded $r$-linear form: $ \Xi\colon \H_1\times \cdots \times\H_r\rightarrow \K$
satisfying:
$$
\sum_{i_1,\ldots,i_r\geq1} |\Xi(e^{(1)}_{i_1},\ldots,e^{(r)}_{i_r})|^2 <\infty,
$$
there exists a unique bounded linear form $L$ on $\T$ so that $\Xi=L\circ\Psi$.
This universal property allows to identify $\Xi$ to an element of $\T^*$.
\vskip2mm
\begin{proposition}\label{pro:cr}
For $F,G\in \F_{HS}$ and $r\geq1$, the map $(x,\eta)\mapsto C_r(F,G)(x,\eta)$ belongs to the
space of functions of Hilbert-Schmidt type $\F_{HS}$.
\end{proposition}

\begin{proof}
Each term in the finite sum~(\ref{eq:cr}) is of the form
\begin{equation}\label{eq:pm}
 \langle\langle D^{(r)}_{\alpha_1\cdots \alpha_r}F(x,\eta),
D^{(r)}_{\alpha^\flat_1\cdots \alpha^\flat_r}G(x,\eta)\rangle\rangle,
\end{equation}
where $\alpha_1,\ldots, \alpha_r= 1$ or $2$. From the definition of $\F_{HS}$, expression~(\ref{eq:pm})
is well defined for any $F,G\in \F_{HS}$ and thus defines a function  on $W$:
$$
\Phi\colon(x,\eta)\mapsto \langle\langle D^{(r)}_{\alpha_1\cdots \alpha_r}F(x,\eta),
D^{(r)}_{\alpha^\flat_1\cdots \alpha^\flat_r}G(x,\eta)\rangle\rangle.
$$

The case $r=1$ has been already proved in Prop.~\ref{pro:poihs}. For $r\geq2$, we need
to slightly modify the argument used in the proof of Prop.~\ref{pro:poihs}
since the bilinear map $\langle\langle\ ,\ \rangle\rangle$ defined by~(\ref{conr}) is not
a bounded bilinear form on the product of Banach spaces:
\begin{equation}\label{pairun}
\mathcal{L}^r(\H^{(\alpha_1)},\ldots,\H^{(\alpha_r)};\K)\times
\mathcal{L}^r(\H^{(\alpha^\flat_1)},\ldots, \H^{(\alpha^\flat_r)};\K).
\end{equation}

In order to show that $\Phi$ is in $C^\infty(W,\K)$, we shall use the universal property
of  the Hilbert tensor product $\otimes$ mentioned above. For $F,G\in \F_{HS}$,
we can consider the bounded $r$-linear maps
 $ D^{(r)}_{\alpha_1\cdots \alpha_r}F(x,\eta)$ and $D^{(r)}_{\alpha^\flat_1\cdots \alpha^\flat_r}G(x,\eta)$
as bounded linear forms on  $\H^{(\alpha)}\equiv\H^{(\alpha_1)}\otimes\cdots\otimes \H^{(\alpha_r)}$
and $\H^{(\alpha^\flat)}\equiv\H^{(\alpha^\flat_1)}\otimes\cdots\otimes \H^{(\alpha^\flat_r)}$, respectively.
Then the unbounded bilinear map $\langle\langle\ ,\ \rangle\rangle$ on the product of spaces~(\ref{pairun})
restricts to the natural pairing of $\H^{(\alpha^\flat)}\sim \H^{(\alpha)*}$ and $\H^{(\alpha)}$
which is a smooth map. This shows that $\Psi$ belongs to $C^\infty(W,\K)$.

An argument similar to the one used in the proof of Prop.~3 shows that the partial derivatives
of $\Phi$ involve a finite sum of terms of the form:
$$
\sum_{i_1,\ldots i_r\geq1} \partial_{i^{(\alpha_1)}_1\cdots i^{(\alpha_r)}_r j_1^\sharp\cdots j_a^\sharp}F(x,\eta)
\ \partial_{i^{(\alpha^\flat_1)}_1\cdots i^{(\alpha^\flat_r)}_r k_1^\sharp\cdots k_b^\sharp}G(x,\eta),
$$
where we have used the notations introduced at the beginning of Subsection~3.2.
A~direct application of the Cauchy-Schwarz inequality gives:
\begin{align*}
&\sum_{j_1,\ldots, j_a\geq1}
\sum_{k_1,\ldots, k_b\geq1}
\Big|
\sum_{i_1,\ldots i_r\geq1} \partial_{i^{(\alpha_1)}_1\cdots i^{(\alpha_r)}_r j_1^\sharp\cdots j_a^\sharp}F
\ \partial_{i^{(\alpha^\flat_1)}_1\cdots i^{(\alpha^\flat_r)}_r k_1^\sharp\cdots k_b^\sharp}G
\Big|^2 \leq\\
& \sum_{i_1,\ldots i_r,j_1,\ldots, j_a\geq1}
\big|\partial_{i^{(\alpha_1)}_1\cdots i^{(\alpha_r)}_r j_1^\sharp\cdots j_a^\sharp}F\big|^2\
 \sum_{i_1,\ldots i_r,k_1,\ldots, k_b\geq1}
\big| \partial_{i^{(\alpha^\flat_1)}_1\cdots i^{(\alpha^\flat_r)}_r k_1^\sharp\cdots k_b^\sharp}G\big|^2<\infty.
\end{align*}
This shows that  $\Phi$ is of Hilbert-Schmidt type and hence $(x,\eta)\mapsto C_r(F,G)(x,\eta)$
belongs to $\F_{HS}$.\hfill$\square$
\end{proof}
\vskip2mm
We summarize all the previous facts in the following:
\vskip2mm
\begin{theorem}
Let the $C_r$'s be given by (\ref{eq:cr}), then the formula
\begin{equation}
F\star^M_\hbar G = F G + \sum_{r\geq1}\frac{\hbar^r}{r!} C_r(F,G),
\end{equation}
defines an associative product on $\F_{HS}[[\hbar]]$ and hence a
 deformation quantization $(W,\F_{HS}[[\hbar]],\star^M_\hbar)$
of the Poisson space $(W,\F_{HS},\{\cdot,\cdot\})$.
\end{theorem}

\begin{proof}
That $\star^M_\hbar\colon\F_{HS}[[\hbar]]\times \F_{HS}[[\hbar]]\rightarrow \F_{HS}[[\hbar]]$
is a bilinear map is a direct consequence of Prop.~\ref{pro:cr} and Prop.~\ref{pro:subhs}.
Associativity follows from the same combinatorics  used in the finite-dimensional case and the
fact that the derivatives distribute in the pairing $\langle\langle\ ,\ \rangle\rangle$
defining the $C_r$'s.\hfill$\square$
\end{proof}

\section{On the equivalence of deformation quantizations on $W$}

We end this article by a discussion on the issue of equivalence of star-products on $W$.
In the finite-dimensional case (i.e. on $\R^{2n}$ endowed with its canonical Poisson bracket),
it is well known that all star-products are equivalent to each other. The situation we are dealing
with here, although it is a direct generalization of the flat finite-dimensional case, allows
inequivalent deformation quantizations. We will illustrate this fact on a family
of star-products of exponential type containing the important case of the normal star-product.

\subsection{The Hochschild complex}

The space of functions of Hilbert-Schmidt type $\F_{HS}$  being an associative
algebra over $\K$, we can consider the Hochschild complex
$C^\bullet(\F_{HS},\F_{HS})$ and its cohomology $H^\bullet(\F_{HS},\F_{HS})$.

One has first to specify a class of cochains that would define the Hochschild complex.
Here the cochains are simply  $r$-differential operators in the sense of Def.~\ref{def:op}
which vanishes on constants. The case where $\F=\F_{HS}$ in Def.~\ref{def:op}
allows a more precise description of the $r$-differential operators. Consider
an $r$-differential operator defined by:
$$
\mathbf{A}(F_1,\ldots,F_r)(w)=\sum_{n_1,\ldots,n_r\geq 0}
a^{(n_1,\cdots ,n_r)}(w).(D^{(n_1)}F_1(w),\cdots,D^{(n_r)}F_r(w)),
$$
where $F_1,\ldots,F_r\in \F_{HS}$ and $w=(x,\eta)\in W$.
For $F\in \F_{HS}$, the higher derivative $D^{(m)}F(x,\eta)$ defines an element
of the $m^{\rm th}$ tensor power of $W$ (here we identify $W^*$ with $W$).
For a fixed $w=(x,\eta)\in W$ and $m\geq0$, the linear map $F\mapsto D^{(m)}F(x,\eta)$ from
$\F_{HS}$ to $W^{\stackrel{m}{\otimes}}$ is onto and we can look at the restriction of the
$r$-linear form
$$
a^{(n_1,\ldots,n_r)}(w)\colon \mathcal{D}_w^{(n_1,\ldots,n_r)}\subset
\mathcal{L}^{n_1}_{\rm sym}(W,\K)\times\cdots\times \mathcal{L}^{n_r}_{\rm sym}(W,\K)\rightarrow \K,
$$
to the product $W^{\stackrel{n_1}{\otimes}}\times\cdots\times W^{\stackrel{n_r}{\otimes}}$
as a bounded $r$-linear form:
$$
\tilde a^{(n_1,\ldots,n_r)}(w)\colon W^{\stackrel{n_1}{\otimes}}\times\cdots\times W^{\stackrel{n_r}{\otimes}}
\rightarrow \K,
$$
such that $w\mapsto \tilde a^{(n_1,\ldots,n_r)}(w)$ is a smooth map from $W$ to
$\mathcal{L}^{r}(W^{\stackrel{n_1}{\otimes}}\times\cdots\times W^{\stackrel{n_r}{\otimes}},\K)$.

The Leibniz rule for the derivatives of a product can be written here for $F,G\in  \F_{HS}$ as:
\begin{align*}
D^{(m)}&(FG)(w).(w_1,\ldots,w_m) \cr
&= \frac{1}{m!} \sum_{\sigma\in \mathfrak{S}_m} \sum_{k=0}^{k=m}
\binom{m}{k}\ (D^{(k)}F(w) \otimes D^{(m-k)}G(w)) (w_{\sigma_1}\otimes\cdots\otimes w_{\sigma_m}),
\end{align*}
where $\mathfrak{S}_m$ is the symmetric group of degree~$m$.

Let
$\mathbf{A}(F)=\sum_{m\geq0} \tilde a^{(m)}(w).(D^{(m)}F(w))$
be a differential operator with
$\tilde a^{(m)}\in C^\infty(W,\mathcal{L}(W^{\stackrel{m}{\otimes}},\K))$,
then it follows from the above form for the Leibniz rule that for $F,G\in \F_{HS}$,
$\mathbf{A}(FG)$ can be written as a finite sum
$$
\sum_{m_1,m_2\geq0} \tilde b^{(m_1,m_2)}(w).(D^{(m_1)}F(w),D^{(m_2)}G(w))
$$ for some
$\tilde b^{(m_1,m_2)}\in C^\infty(W,\mathcal{L}^2(W^{\stackrel{m_1}{\otimes}}\times W^{\stackrel{m_2}{\otimes}},\K))$,
and thus $(F,G)\mapsto \mathbf{A}(FG)$ is a bidifferential operator. The generalization to $r$-differential operators
of this fact is straightforward.

The Hochschild complex consists of multidifferential operators
vanishing on constants,
i.e., $C^\bullet(\F_{HS},\F_{HS})=\oplus_{k\geq0} C^k(\F_{HS},\F_{HS})$
where, for $k\geq1$, the space of \mbox{$k$-cochains} is:
\begin{align*}
&C^k(\F_{HS},\F_{HS})\\
&=\{\mathbf{A}\colon \F_{HS}^{k}\rightarrow \F_{HS}\ | \
\mathbf{A}\ {\rm is\ a}\ k{\rm -differential\ operator\ vanishing\ on}\ \K\}.
\end{align*}

The differential of a $k$-cochain $\mathbf{A}$ is the  $(k+1)$-linear map  $\delta\mathbf{A}$ given by:
\begin{eqnarray} \label{bH}
\delta\mathbf{A}(F_0,\ldots,F_k)&=&F_0\mathbf{A}(F_1,\ldots,F_k)-\mathbf{A}(F_0F_1,F_2,\ldots,F_k)+
\cdots\\
&+&(-1)^k\mathbf{A}(F_0,F_1,\ldots,F_{k-1}F_k)+(-1)^{k+1}\mathbf{A}(F_0,\ldots,
F_{k-1})F_k.\nonumber
\end{eqnarray}
satisfies $\delta^2=0$, and according to the discussion above
$\delta\mathbf{A}$ is a $(k+1)$-differential operator,
vanishing on constants whenever $\mathbf{A}$ does. Thus $\delta\mathbf{A}$
is a cochain that belongs to $C^{k+1}(\F_{HS},\F_{HS})$,
hence we indeed have a complex.

A $k$-cochain $\mathbf{A}$ is a $k$-cocycle if $\delta\mathbf{A}=0$.
We denote by $Z^k(\F_{HS},\F_{HS})$ the space of $k$-cocycles and by $B^k(\F_{HS},\F_{HS})$ the
space of those $k$-cocycles which are coboundaries.
The $k^{\rm th}$ Hochschild cohomology space of $\F_{HS}$ valued in $\F_{HS}$
is defined as the quotient $H^k(\F_{HS},\F_{HS})=Z^k(\F_{HS},\F_{HS})/B^k(\F_{HS},\F_{HS})$.

\subsection{Star-products of the exponential type}

Let $\mathcal{B}(\H)$ denote the algebra of bounded operators on $\H$ and $\mathcal{B}_2(\H)$, the
two-sided $*$-ideal of Hilbert-Schmidt operators on $\H$ .
We shall describe a family of deformation quantizations
$\{(W,\F_{HS}[[\hbar]],\star^A_\hbar)\}_{A\in \mathcal{B}(H)}$.
Each star-product~$\star^A_\hbar$ where $A\in \mathcal{B}(H)$ shall
be the exponential of a Hochschild $2$-cocycle, with the Moyal star-product
corresponding to the case $A=0$. It will turn out that the set of
 equivalence classes of star-products of this type is parameterized by
$\mathcal{B}(\H)/\mathcal{B}_2(\H)$.

Let $A \in \mathcal{B}(\H)$. For $F,G\in \F_{HS}$, the map
\begin{equation}\label{eadef}
(x,\eta)\mapsto E_A(F,G)(x,\eta)\equiv\langle D_1F(x,\eta), A D_2G(x,\eta)\rangle
+ \langle D_1G(x,\eta), A D_2F(x,\eta)\rangle
\end{equation}
defines a smooth function on $W$ and is symmetric in $F$ and $G$. Moreover we have:
\begin{proposition}\label{lemma:ea}
The bilinear map $(F,G)\mapsto  E_A(F,G)$ is  a Hochschild $2$-cocycle.
\end{proposition}

\begin{proof}
It is clear that $\delta E_A=0$ and, since $E_A$ vanishes on constants, it is sufficient to check
that $E_A$ is a bidifferential operator on $(W,\F_{HS})$, i.e., that the smooth function
$(x,\eta)\mapsto E_A(F,G)(x,\eta)$ is of Hilbert-Schmidt type for any $F,G\in \F_{HS}$.
Let $\{e_i\}_{i\geq 1}$ be an orthonormal basis of $\H$ and $\{e^*_i\}_{i\geq 1}$ the dual basis.
Consider the basis $\{f_i\}_{i\geq 1}$ in $W=\mathcal{H}\oplus \mathcal{H}^*$ defined by
$f_i=(e_{\frac{i+1}{2}},0)$ if $i$ is odd, and $f_i=(0,e^*_{\frac{i}{2}})$ if $i$ is even.
To show that  $E_A(F,G)$ is in $\F_{HS}$ is equivalent to show that
\begin{equation}\label{eahs}
\sum_{i_1,\ldots,i_n\geq1}| D^{(n)}E_A(F,G)(x,\eta).(f_{i_1},\ldots,f_{i_n})|^2 <\infty.
\end{equation}
holds for any $n\geq1$ and $w=(x,\eta)\in W$.

The chain rule applied to
\begin{equation*}
\begin{CD}
\Phi\colon W @>{\Psi_1}>>\H^*\times \H  @>{\Psi_2}>>\K
\end{CD}
\end{equation*}

$$
(x,\eta)\mapsto (D_1F(x,\eta),D_2G(x,\eta)) \mapsto \langle D_1F(x,\eta),AD_2G(x,\eta)\rangle,
$$
\vskip2mm
\noindent shows that the derivatives of $E_A$ distributes in the pairing $\langle\ , \ \rangle$.

The $n^{\rm th }$ derivative of $E_A(F,G)$ in the direction of $(f_{i_1},\ldots,f_{i_n})$ is
a finite sum of terms of the form:
\begin{equation}\label{da}
\langle D^{(r)}D_1F(x,\eta).(f_{k_1},\ldots,f_{k_r}), A D^{(s)}D_2G(x,\eta).(f_{l_1},\ldots,f_{l_s})\rangle,
\end{equation}
where $r+s=n$ and $(k_1,\ldots,k_r, l_1,\ldots,l_s)$ is a permutation of $(i_1,\ldots,i_n)$, and
similar terms  with $F$ and $G$ inverted.
It is worth noting that $D^{(r)}D_1F(x,\eta).(f_{k_1},\ldots,f_{k_r})$ is the element of $\H^*$ defined
by the bounded linear form
$$h\mapsto D^{(r+1)}F(x,\eta).((h,0),f_{k_1},\ldots,f_{k_r})$$
 on $\H$
and, similarly,
$D^{(s)}D_2G(x,\eta).(f_{l_1},\ldots,f_{l_s})$ is the element of $\H$ defined by the bounded
linear form on $\H^*$:
$$
\xi\mapsto D^{(s+1)}G(x,\eta).((0,\xi),f_{l_1},\ldots,f_{l_s}).$$
The modulus squared of (\ref{da}) is bounded by the constant
$
||A||^2 \alpha_{k_1\ldots k_r} \beta_{l_1\ldots l_s}
$, where
\begin{align*}
\alpha_{k_1\ldots k_r}=&\sum_{i\geq1}|D^{(r+1)}F(x,\eta).((e_i,0),f_{k_1},\ldots,f_{k_r})|^2,\\
\beta_{l_1\ldots l_s}=&\sum_{j\geq1}|D^{(s+1)}G(x,\eta).((0,e^*_j),f_{l_1},\ldots,f_{l_s})|^2.
\end{align*}
Since $F,G\in\F_{HS}$, we have $\ds\sum_{k_1,\ldots,k_r\geq1}\alpha_{k_1\ldots k_r}<\infty$ and
$\ds\sum_{l_1,\ldots,l_s\geq1}\beta_{l_1\ldots l_s}<\infty$, from which inequality~(\ref{eahs})
follows.\hfill$\square$
\end{proof}

For any $A \in \mathcal{B}(\H)$, let us define:
\begin{align}\label{coneafg}
C^A_1(F,G) & = \{F,G\} + E_A(F,G)\\
           & = \langle D_1F,(A + \I)D_2G\rangle + \langle D_1G,(A - \I)D_2F\rangle\nonumber,
\end{align}
where $\I$ is the identity operator on $\H$. Plainly, $C^A_1$ is a $2$-cocycle with constant coefficients,
and one can define a star-product by taking the exponential of $C^A_1$:
$$
F\star^A_\hbar G = \exp(\hbar C^A_1)(F,G)= FG + \sum_{r\geq 1} \frac{\hbar^r}{r!}  C^A_r(F,G),
$$
where $C^A_r= (C^A_1)^r$ in the sense of bidifferential operators. This formula
defines an associative product on $\F_{HS}[[\hbar]]$, and
we get a  family of deformation quantizations
$\{(W,\F_{HS}[[\hbar]],\star^A_\hbar)\}_{A\in \mathcal{B}(H)}$ of $(W,\F_{HS},\{,\cdot,\cdot\})$.
This family of star-products is easily described by their symbols.
Let us consider the following family of smooth functions on $W$:
\begin{equation}\label{phiyxi}
\Phi_{y,\xi}(x,\eta) =\exp(\langle\eta,y \rangle + \langle\xi,x \rangle), \quad
x,y\in \H,\ \eta,\xi\in \H^*.
\end{equation}
The $\Phi_{y,\xi}$'s belong to $\F_{HS}$ and from (\ref{coneafg}) we deduce that:
$$
C^A_r(\Phi_{y,\xi},\Phi_{y',\xi'}) =
(\langle\xi, (A + \I)y' \rangle + \langle\xi',(A - \I)y \rangle)^r\ \Phi_{y+y',\xi+\xi'},
$$
and consequently:
\begin{equation}\label{asymbol}
\Phi_{y,\xi}\star^A_\hbar \Phi_{y',\xi'} =
\exp\big(\hbar(\langle\xi, (A + \I)y' \rangle + \langle\xi',(A - \I)y \rangle)\big)\ \Phi_{y+y',\xi+\xi'}.
\end{equation}

\begin{example}
Set $A=\I$ in (\ref{coneafg}), then $C^\I_1(F,G) = 2 \langle D_1F,D_2G\rangle$ and
the corresponding star-product reads
\begin{equation}\label{normal}
F\star^\I_\hbar G= F G+\sum_{r\geq1}\frac{(2\hbar)^r}{r!}
\langle D^{(r)}_{1\cdots 1}F,D^{(r)}_{2\cdots 2}G\rangle.
\end{equation}
It is the well-known normal star-product
(or Wick or standard depending on the interpretation of the variables $(x,\eta)\in W$).
The cochains defining the normal star-product $C^\I_r$ correspond to only one term
in the sum defining the $r^{\rm th}$ cochain of the Moyal star-product,
namely the term corresponding
to $\alpha_1=\cdots=\alpha_r=1$ and $\beta_1=\cdots=\beta_r=2$
in the sum  (\ref{eq:cr}). One would expect that conditions~(\ref{hs}) defining
the functions of Hilbert-Schmidt type are not all needed in order to make sense
of the normal star-product and would guess that this product can be defined
on a wider class of functions.
Actually, the normal star-product defines a deformation quantization on a larger
space of functions (containing the free Hamiltonian) that we shall describe
in a forthcoming paper.
\end{example}

At this stage, a natural question arises: are the deformation quantizations
$\{(W,\F_{HS}[[\hbar]],\star^A_\hbar)\}_{A\in \mathcal{B}(H)}$
equivalent to each other? The answer is given in the:

\begin{proposition}\label{coboundary}
Let $A \in \mathcal{B}(\H)$. The Hochschild $2$-cocycle $E_A$ defined by (\ref{eadef}) is a coboundary
if and only if $A$ is a Hilbert-Schmidt operator.
\end{proposition}
\begin{proof}
Let $A$ be a Hilbert-Schmidt operator on $\H$. The map from $W$ to $\K$ defined by
$(x,\eta)\mapsto \langle\eta,A x \rangle$ is of Hilbert-Schmidt type and therefore
there exists a bounded linear form $\tilde A\colon \H^*\otimes\H\rightarrow \K$
so that
$\langle\eta,A x \rangle = \langle\tilde A, \eta\otimes x \rangle$.
The $2$-cocycle $E_A$ can then be written as:
$$
E_A(F,G)= \langle\tilde A, D_1F\otimes D_2G + D_1G\otimes D_2F \rangle,\quad F,G\in \F_{HS}.
$$
For $F \in \F_{HS}$, the mixed derivative $D^{(2)}_{12}F$ belongs to
$C^\infty(W, \mathcal{L}(\H \otimes \H^*,\K))\sim C^\infty(W, \H^* \otimes \H)$ and
$T_A(F)(x,\eta) = - \langle\tilde A, D^{(2)}_{12}F(x,\eta) \rangle$ defines a differential
operator on $(W, \F_{HS})$ vanishing on constants. For any two functions in $\F_{HS}$,
the Leibniz rule reads:
$$
D^{(2)}_{12}(FG)
= F D^{(2)}_{12}(G) + G D^{(2)}_{12}(F) + D_1F\otimes D_2G + D_1G\otimes D_2F,
$$
it belongs to $C^\infty(W, \H^* \otimes \H)$
and a simple computation gives $\delta T_A= E_A$, therefore $E_A$ is a coboundary if $A$
is in the Hilbert-Schmidt class.

Conversely, if $E_A$ is a coboundary, there exists a differential operator $S$
on $(W, \F_{HS})$ vanishing on constants, so that $E_A=\delta S$. If a term
of degree one occurs in $S$ (a derivation) it can be subtracted without
changing $\delta S$, hence we can assume that $S$ has the form:
$$
S(F)(x,\eta)= \sum_{m\geq2} a^{(m)}(x,\eta).(D^{(m)}F(x,\eta)),
$$
where $a^{(m)}\in C^\infty(W, \mathcal{L}(W^{\stackrel{m}{\otimes}},\K))$ and
only finitely many of them are nonzero. By computing $\delta S$ and using
$E_A=\delta S$, we find that only the term of degree $2$ contributes:
\begin{align*}
\langle & D_1F(w)\,A D_2G(w)\rangle +\langle D_1G(w),A D_2F(w) \rangle\\
&=- a^{(2)}(w).(DF(w)\otimes DG(w) + DG(w)\otimes DF(w)),\quad w=(x,\eta)\in W.
\end{align*}
If we evaluate the equality above on $F(x,\eta)=\langle \xi,x\rangle$, $\xi\in \H^*$,
and $G(x,\eta)=\langle \eta,y\rangle$, $y\in \H$, we find (with a slight abuse of notations):
$$
\langle \xi,A y\rangle = - a^{(2)}(w).(\xi\otimes y),\quad \forall \xi\in\H^*, \forall y\in\H,
$$
from which follows that $A\in \mathcal{B}(\H)$ is a Hilbert-Schmidt operator on $\H$,
as $a^{(2)}(w)$ is a bounded linear form on $\H^*\otimes \H$.\hfill$\square$
\end{proof}

As an  immediate consequence of Prop.~(\ref{coboundary}), we deduce a classification result
for deformation quantizations of exponential type
$\{(W,\F_{HS}[[\hbar]],\star^A_\hbar)\}_{A\in \mathcal{B}(H)}$.

\begin{theorem}
Let $A,B\in \mathcal{B}(H)$. Two deformation quantizations $(W,\F_{HS}[[\hbar]],\star^A_\hbar)$
and $(W,\F_{HS}[[\hbar]],\star^B_\hbar)$
are equivalent if and only if $A-B$ is in the Hilbert-Schmidt class. Consequently,
the set of equivalence classes of $\{(W,\F_{HS}[[\hbar]],\star^A_\hbar)\}_{A\in \mathcal{B}(H)}$
is parameterized by $\mathcal{B}(\H)/\mathcal{B}_2(\H)$.
\end{theorem}
\begin{proof}
Suppose that $\star^A_\hbar$ and $\star^B_\hbar$ are equivalent, i.e., there exists
a formal series of differential operators vanishing on constants:
$T=\mathrm{Id}_{\F_{HS}} + \sum_{r\geq1} \hbar^r T_r$, so that $T(F\star^A_\hbar G)= TF \star^B_\hbar TG$.
Then it follows that $C_1^A=C_1^B + \delta T_1$  and, from the definitions (\ref{eadef}) and (\ref{coneafg})
of $E_A$ and $C_1^A$, we have $E_{A-B}= E_A - E_B = \delta T_1$, showing that $E_{A-B}$ is a coboundary
and hence $A-B$ is a Hilbert-Schmidt operator on $\H$.

Conversely, if $S\equiv A-B$ is a Hilbert-Schmidt operator, it defines a bounded
linear form $\tilde S$  on $\H^*\otimes \H$ and a differential operator
$T_1(F) = - \langle\tilde S, D^{(2)}_{12}F \rangle$ on
$(W,\F_{HS})$ (cf. the proof of Prop.~\ref{coboundary}).
Since the star-products $\star^A_\hbar$ and $\star^B_\hbar$ are defined by
constant coefficient bidifferential operators, it is sufficient to establish
the equivalence at the level of symbols.
Now define the formal series of differential operators $T=\exp(\hbar T_1)$.
Its symbol is given by
$T(\Phi_{y,\xi})= \exp(\hbar \langle \xi,(B-A)y \rangle)\Phi_{y,\xi}$,
where $\Phi_{y,\xi}$ has been defined in (\ref{phiyxi}).
Using the symbol~(\ref{asymbol}) associated to a star-product, we find:
$$
T(\Phi_{y,\xi}\star^A_\hbar \Phi_{y',\xi'})=T\Phi_{y,\xi} \star^B_\hbar T\Phi_{y',\xi'},
\quad y,y'\in \H, \xi,\xi'\in \H^*.
$$
Therefore the deformation quantizations $(W,\F_{HS}[[\hbar]],\star^A_\hbar)$ and
$(W,\F_{HS}[[\hbar]],\star^B_\hbar)$ are equivalent.\hfill$\square$
\end{proof}

The Moyal and normal star-products correspond to $A=0$ and  $A=\I$ in (\ref{coneafg}), respectively.
Since the identity operator on the infinite-dimensional Hilbert space $\H$ is not in the
Hilbert-Schmidt class, we have:
\begin{corollary}
The Moyal and normal star-products are not equivalent deformations on $(W,\F_{HS}[[\hbar]])$.
\end{corollary}

\begin{remark}
One can generalize the class of exponential type of star-products
by allowing formal series with coefficients in $\mathcal{B}{(\H)}$ in (\ref{coneafg}) or
(\ref{asymbol}). The set of equivalence classes would then be
$(\mathcal{B}(\H)/\mathcal{B}_2(\H))[[\hbar]]$. Since we did not show that any star-product
on $(W,\F_{HS},\{\cdot,\cdot\})$ is equivalent to a star-product of the exponential type,
$(\mathcal{B}(\H)/\mathcal{B}_2(\H))[[\hbar]]$ can only be considered as a
 \textit{lower bound} for the classification space of all the star-products on
$(W,\F_{HS},\{\cdot,\cdot\})$.
\end{remark}

\begin{remark}
Recall that all the star-products on $\R^{2n}$ endowed with its canonical Poisson
structure are equivalent to each other. This fact should be put in relation with Von Neumann's
uniqueness theorem on the irreducible (continuous) representations of
Weyl systems associated to the canonical  commutation relations (CCR) $\{q_i,p_j\}=\delta_{ij}$.
The inequivalent representations of Weyl systems associated to
the infinite dimensional CCR have been described long time ago
by G\aa rding and Wightman \cite{GW}, and Segal \cite{Se}. Here the existence
of inequivalent deformation quantizations conveys the idea that there should be
a close link  between the set of equivalence classes of star-products and
representations of Weyl systems associated to the CCR. It might turn out that they
actually are identical.
\end{remark}

\section*{Acknowledgments}
The author would like to thank Y. Maeda and S. Watamura, the organizers of the wonderful
meeting  \textit{International Workshop on Quantum Field Theory
and Noncommutative Geometry} held in Sendai (November 2002), for their warmest hospitality.

config: arxiv


TITLE: How to derive relation between relative and log deviations?
QUESTION [0 upvotes]: I ran into the following approximation:
$$\frac{x_t - x_0}{x_0} \approx \log(x_t/x_0) + \frac{1}{2}\log(x_t/x_0)^2 $$
which is supposedly a second-order approximation around a fixed value $x_0$. How does this come about?

REPLY [1 votes]: Taylor expand $\ln(x/x_0)$ around $x_0$
$$
\ln\left(\frac{x}{x_0}\right) = \frac{x - x_0}{x_0} - \frac{(x - x_0)^2}{2x_0^2} + \frac{(x - x_0)^3}{3x_0^3} + \cdots
$$
Similarly
$$
\ln^2\left(\frac{x}{x_0}\right) = \frac{(x - x_0)^2}{x_0^2} - \frac{(x - x_0)^3}{x_0^3} + \cdots
$$
So that
$$
\ln\left(\frac{x}{x_0}\right) + \frac{1}{2}\ln^2\left(\frac{x}{x_0}\right) = \frac{x-x_0}{x_0} - \frac{(x - x_0)^3}{6x_0^3} + \cdots
$$


TITLE: Finding the Euler angle/axis from a 2 axes rotation but that lies on the original 2 axes' plane
QUESTION [1 upvotes]: If I rotate the XY plane about the y-axis by an angle ψ and then intrinsically rotate it about the new plane's x-axis by an angle φ, the axis created by the intersection of the original XY plane and the final plane will be a Euler axis/angle.
What are the formulae to obtain the axis angle (relative to the x or y axis) and the angle of rotation?
This question has it's roots in a machining problem I have where the parts can only be rotated on a single axis and the design has 2 rotations.
I have tried several different solutions given both here and on Wikipedia using Euler and Rodrigues rotation formulae but they all place the rotation axis in 3D space. I need it to lie on the XY plane but I can't work out how to zero out the z-axis component.

REPLY [0 votes]: SOLUTION FORMULAE
For rotations about the x-, y- and z-axes with angles $ \phi $, $ \theta $ and $ \psi $ respectively, the Euler axis of rotation that lies on the XY plane and it's associated rotation angle can be derived from the following formulae using trig functions only.
The Euler axis of rotation is offset from  the y-axis by an angle  $ \alpha $  about the z-axis where:
$$
\alpha =  \tan^{-1} \biggl[(- \sin \phi \times \cos \psi + \cos \phi \times \sin \theta \times \sin \psi )\\ \div ( \sin \phi \times \sin \psi + \cos \phi \times \sin \theta \times \cos \psi) \biggr]\\
$$
and the rotation angle $ \beta $ about the Euler axis of rotation and offset from the XY plane is given by
$$
\\
\beta =  \pi /2 - \sin^{-1} \left[ \sqrt{ ( \sin \phi \times \sin \psi + \cos \phi \times \sin \theta \times \cos \psi )^{2}\\ + (-\sin \phi \times \cos \psi + \cos \phi \times \sin \theta \times \sin \psi)^{2} } \right]
\\
$$
In the case where only 2 rotations exist with $ \psi = 0 $ we substitute for $ \cos \psi = 1 $ and $ \sin \psi = 0 $.
The Euler angles $ \alpha $ and $ \beta $ then simplify to:
\begin{align*}
\alpha & = tan^{-1} \biggl[ - \sin  \phi  / (  \cos  \phi  \times  \sin  \theta ) \biggr]
\\
\\
\beta & = \pi /2 - \sin^{-1} \biggl[ \sqrt{(\cos \phi \times \sin \theta )^{2} + (-\sin \phi)^{2}  } \biggr]
\\
\end{align*}
All formulae are derived from the $A_{31}$ and $A_{32}$ elements of the rotation matrix expressed in terms of the sine and cosine of the rotation angles (i.e. the x and y components of the new z-axis).
In the above case this is the "new" z-axis relative to the original XY plane and is coincident with the origin which serves as the centre point for all rotations, but it may be any point on any plane.


TITLE: Euler series transformation
QUESTION [0 upvotes]: Consider an infinite power series
\begin{equation}
S(x) = \sum\limits_{n=0}^\infty b_n x^n
\end{equation}
If the coefficients of that series can be written as an $n$th backwards difference of some other set of coefficients then the series can be resummed. In other words the following identity holds:
\begin{equation}
S(x) = \frac{1}{1+x} \sum\limits_{k=0}^\infty a_k \left(\frac{x}{1+x}\right)^k
\end{equation}
where $\Delta^n a_n := \sum\limits_{k=0}^n \binom{n}{k} (-1)^{n-k} a_k = b_n$
or $a_n = \sum\limits_{k=0}^n \binom{n}{k} b_k$.
Now, let us assume that
\begin{equation}
a_n = \frac{H^{(m_1)}_{n+1}}{(n+1)^m}
\end{equation}
where $m$ and $m_1$ are non-negative integers and $H^{(m_1)}_{n+1}$ are generalized Harmonic numbers.
How do I compute the coefficients $\left\{b_n\right\}_{n=0}^\infty$ in this case?
Hint: In Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$ the answer is found for the case $(m,m_1)=(1,2)$.

REPLY [1 votes]: Let us consider the case $m_1=0$ at first. In other words we take:
\begin{equation}
a^{(m)}_n = \frac{1}{(n+1)^m}
\end{equation}
where $m \ge 1$ is a fixed integer. By generalizing the methods from the question cited above (or by using the trick $1/(k+1)^m = \int\limits_0^\infty \theta^{m-1}/(m-1)! \exp[-\theta (k+1)] d\theta$ we easily get:
\begin{equation}
b^{(m)}_n = \frac{(-1)^n}{n+1} \frac{S_{m-1}(n)}{(m-1)!}
\end{equation}
where
\begin{equation}
S_m(n) = \left\{
\begin{array}{rr}
H_{n+1}                   & \mbox{if $m=1$}\\
H_{n+1}^2 + H^{(2)}_{n+1} & \mbox{if $m=2$}\\
H_{n+1}^3 + 3 H_{n+1} H_{n+1}^{(2)}+ 2 H^{(3)}_{n+1} & \mbox{if $m=3$}\\
H_{n+1}^4 + 6 H_{n+1}^2 H_{n+1}^{(2)} +3 (H_{n+1}^{(2)})^2 +8 H_{n+1} H_{n+1}^{(3)}+ 6 H^{(4)}_{n+1} & \mbox{if $m=4$}\\
\vdots
\end{array}
\right.
\end{equation}
and in general:
\begin{equation}
S_m(n) = \sum\limits_{\lambda=1}^m \frac{1}{\lambda!}
\sum\limits_{\begin{array}{r} p_1+\cdots+p_\lambda=m \\ p_1 \ge 1, \cdots,p_\lambda\ge 1\end{array}} \prod\limits_{q=1}^\lambda \frac{H_{n+1}^{(p_q)}}{p_q}
\end{equation}
The Euler series transformation results in this case in a sequence of following identities:
\begin{equation}
\frac{1}{x} Li_m\left( \frac{x}{x+1} \right) = \sum\limits_{n=0}^\infty b^{(m)}_n x^n
\end{equation}
In particular:
\begin{eqnarray}
Li_1\left( \frac{x}{x+1}\right) &=& - \frac{1}{0!} Li_1(-x) \quad \\
Li_2\left( \frac{x}{x+1}\right) &=& - \frac{1}{1!} \sum\limits_{n=1}^\infty \frac{H_n}{n} (-x)^n \\
Li_3\left( \frac{x}{x+1}\right) &=& - \frac{1}{2!}\sum\limits_{n=1}^\infty \frac{H_n^2+H_n^{(2)}}{n} (-x)^n \\
Li_4\left( \frac{x}{x+1}\right) &=& - \frac{1}{3!}\sum\limits_{n=1}^\infty \frac{H_n^3+3 H_n H_n^{(2)}+2 H_n^{(3)}}{n} (-x)^n \\
\vdots
\end{eqnarray}
Now we consider the case $m=0$. In other words we take:
\begin{equation}
a^{(m_1)}_n = H^{(m_1)}_{n+1}
\end{equation}
where $m_1 \ge 1$ is a fixed integer. Using the same methods as above we establish the following:
\begin{equation}
b^{(m_1)}_n = \frac{(-1)^{n-1}}{n(n+1)} \frac{\bar{S}_{m-1}(n)}{(m-1)!}
\end{equation}
where
\begin{equation}
\bar{S}_m(n) = S_m(n) - S_{m-1}(n)
\end{equation}
This leads to following identities:
\begin{equation}
\frac{x+1}{x} Li_m\left( \frac{x}{x+1} \right) = \sum\limits_{n=0}^\infty b^{(m_1)}_n x^n
\end{equation}
In particular:
\begin{eqnarray}
 Li_1\left( \frac{x}{x+1} \right) &=& \log(1+x) \\
Li_2\left(\frac{x}{x+1}\right) &=& -Li_2\left(\frac{1}{x+1}\right)-\log ^2(x+1)+\log (x) \log (x+1)+\frac{\pi ^2}{6} \\
Li_3\left(\frac{x}{x+1}\right) &=&  \frac{x}{x+1}\left(1+\frac{1}{2}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n (n+1)} \left(H_{n+1}^2+H_{n+1}^{(2)}-2 H_{n+1}\right) x^n \right) \\
\vdots
\end{eqnarray}


TITLE: Find the eigenvalues and eigenvectors of the matrix with all diagonal elements as $d$ and rest $1$
QUESTION [7 upvotes]: A matrix has all elements 1 except the diagonal elements. It is an $n\times n$ matrix. What are the eigenvectors and eigenvalues ?
Solving book problems in Strang book and stuck on this one and I have no idea where to begin ?

REPLY [0 votes]: The value of your eigenvalues and diagonal elements are the same!