User tom - MathOverflow

Answer by Tom for Solve for $A$ and $B$ in $AXB=Y$

2013-05-09T14:16:56Z

This should be a comment but I haven't got enough rep, sorry. I don't know how you want to apply the result. So, I'm wondering whether a linear polynomial whose coefficients are $m\times n$ and $n \times m$ matrices would be sufficient for your application. This can be easily achieved by using elementary matrices in order to extract $X$'s entries.

EDIT for elaboration

Let $E_{ij}=E_{ij}^{(n)}$ denote the $n\times n$ matrix that has got zero entries everywhere except for the i-th row and j-th column, i.e. $ \left( E_{ij} \right)_{kl}= \delta _{ik} \delta _{jl} $ .

Then $ E_{ii}\cdot X \cdot E_{jj} $ equals $x_{ij}E_{ij}$ where $x_{ij}= (X)_{ij} $.

Well, the embedding $$\iota\colon M(n,R) \to M(m,R) \quad ; \quad M \mapsto \begin{pmatrix} M & 0 \\ 0 & 0 \end{pmatrix}$$ can be described by the matrix $J=(I_n \ 0_{m-n})$, i.e. $\iota(M)=J^t\cdot M\cdot J$.

Let me just steal the next definition from wikipedia http://en.wikipedia.org/wiki/Elementary_matrices

$$T_{i,j} = \begin{bmatrix} 1 & & & & & & & \\ & \ddots & & & & & & \\ & & 0 & & 1 & & \\ & & & \ddots & & & & \\ & & 1 & & 0 & & \\ & & & & & & \ddots & \\ & & & & & & & 1\end{bmatrix}$$

So $T_{ij}\cdot A$ is the matrix produced by exchanging row $i$ and row $j$ of $A$.

Suppose $ y_{kl} = \sum_{ij} z_{kl}^{ij} \cdot x_{ij} $ where $z_{kl}^{ij}$ lies in $\mathbb Z$ and $y_{kl}=(Y)_{kl}$ then

$$Y=\sum_{ijkl} z_{kl}^{ij} \cdot T_{ik}^{(m)} \cdot J^t \cdot E_{ii}^{(n)} \cdot X \cdot E_{jj}^{(n)} \cdot J \cdot T_{jl}^{(m)} . $$

Or, as I just realized we can permute

$$Y=\sum_{ijkl} z_{kl}^{ij} \cdot T_{ik}^{(m)} \cdot E_{ii}^{(m)}\cdot J^t\cdot X\cdot J \cdot E_{jj}^{(m)} \cdot T_{jl}^{(m)} . $$

But both formulas give the exact same shortened version

$$ Y = \sum_{ijkl} A_{kl}^{ij} \cdot X \cdot B_{l}^{ij} $$

where $B_{kl}^{ij}$ is independent of $k$.

Answer by Tom for Interesting applications (in pure mathematics) of first-year calculus

2013-04-05T12:22:25Z

What I like(d) most is defining an analytic function that describes some number theoretic phenomena. One thing I remember is from Winfried Kohnen's postech lecture http://www.mathi.uni-heidelberg.de/~winfried/siegel2.pdf , see pages 1-3 for more details. He starts with the standard inner product on $\mathbb{R}^m$ viewed as a quadratic form $$Q(x):=x^t x.$$ We are interested in the number $r_Q(t)$ of tuples of squares of inetegers that add up to a natural number $t$, i.e.

$$ r_Q(t):= # \left{ g \in \mathbb{Z}^m : Q(g)=(g_1)^2+ \dots + (g_4)^2=t \right} .$$

They can be computed via this power series

$$ \theta_Q(z) = 1+ \sum_{t\geq 1} r_Q(t)\ \exp(2\pi i tz) $$

that is in fact $\theta_Q$ is a modular form of weight 2 w.r.t. $\Gamma_0$. Therefore (ok here is some kind of black box for the students), its Fourier coefficients can be given by $$r_Q(t)= 8 \left( \sigma_1(t)-4\cdot \sigma_1\left(\frac{t}{4}\right) \right)$$

where $\sigma_k(t)$ denotes the divisor function $$\sigma_k(t):=\sum_{d|t} d^k.$$

While writing this I was wondering whether the prime number theorem and elegant proofs of the fundamental theorem of algebra are too well known.

p.s. sorry for messing up the formulas again.

Answer by Tom for Reference on generators of subgroups of symplectic groups

2013-04-02T15:30:27Z

I just wanted to share a tiny part of the solution with you. The group $\Gamma_{2,0}[2]$ is generated by the matrices $\begin{pmatrix}I_g & S \\ 0_g & I_g \end{pmatrix}$ where $S=S^t$,

$\begin{pmatrix}I_g & 0_g \\ 2 \cdot S & I_g \end{pmatrix}$ where $S=S^t$

and $\begin{pmatrix}U^t & 0_g \\ 0_g & U^{-1} \end{pmatrix}$ where $U \in GL(2,\mathbb{Z})$. The reference is http://arxiv.org/abs/1001.0324 page 6.

Reference on generators of subgroups of symplectic groups

2013-03-05T16:46:13Z

We should start with the definition of the symplectic group for an arbitrary ring $R$. The symplectic group $Sp(g,R)$ is the subgroup of $SL(2g,R)$ such that all elements satisfy $M=J_g^t M J_g$ with $J_g$ being the canonical almost complex structure - or involution whatever you prefer to call it.

A generic element of $Sp(g,R)$ is denoted by $M=(A B \ C D)$ where $A,B,C,D$ are $g\times g$ matrices.

Note bene, if $R$ is Euclidean then $Sp(g,R)$ is generated by the involution $J_g$ and the translations $\begin{pmatrix}I_g\ S \\ 0_g \ I_g \end{pmatrix}$ sending $Z$ to $Z+S$ where $S$ is a symmetric $g \times g$ matrix.

Why am I interested in these generators ?

Well, first of all I am interested in modular forms. They are holomorphic functions $f:\mathbb{H}_g \to V $ transforming under a subgroup $\Gamma$ of a symplectic group as follows $$f(M \cdot Z)=j(M,Z)\cdot f(Z)\quad \quad \forall\ M \in \Gamma ,$$ where $j$ is a factor of automorphy. This means that $j: \Gamma \times \mathbb{H}_g \to GL(V)$ is holomorphic in the second variable and satisfies the cocycle relation $$j(MN,Z)=j(M,N \cdot Z) \cdot j(N,Z).$$

Hence, it suffices to check the first equation only for the generators of $\Gamma$.

Sometimes we have modular forms to a proper subgroup and even know how they transform under the full symplectic group. But they do not transform with a factor of automorphy. Examples would be the theta series $$f_a(Z):=\sum_{\nu \in \mathbb{Z}^g}{exp \left(2\pi i \left(\nu+\frac{a}{2}\right)^t Z \left(\nu+\frac{a}{2}\right) \right)}, \quad \quad a \in \mathbb{F}_2^g$$ They are modular forms to certain proper subgroups. But the action of $Sp(g,\mathbb{Z})$'s generators can be given quite easily, roughly speaking : the translations scale the thetas and involution returns a linear combination of all thetas. That way, it is possible to determine whether a polynomial in the different theta series is a modular form to the full modular group.

The actual question

I would be very pleased if someone could give a reference for the generators of subgroups of $Sp(g,\mathbb{Z})=\Gamma_g$ like $$ \Gamma_g[q]:= ker\left(Sp(g,\mathbb{Z})\to Sp(g,\mathbb{Z}/q\mathbb{Z})\right)$$ $$ \Gamma_{g,0}[N]:=\left\{ M \in \Gamma_g : C \equiv 0 \mod N \right\} $$ $$ \Gamma_{g}^{0}[N]:=\left\{ M \in \Gamma_g : B \equiv 0 \mod N \right\} $$ and others if you know them, too. In particular, I am interested in $g=2$. I guess this way it is faster and I cannot make any mistakes. As hinted above it would be also nice if these generators could be given in terms of the generators of $Sp(g,\mathbb{Z})$.

Thanks Tom

p.s. it would be nice if someone could help me fixing the brackets in the above definition of $\Gamma_{g,0}[N]$.

edit1 : added notation $M=(A B \ C D)$

Are there cusp forms for the full modular group Sp(2,Z) and representations det^3 \otimes Sym^2j(\rho_standard)

2012-07-18T20:11:12Z

What are modular forms or cusps forms, resp. ?

We start with defining their common domains $\mathbb{H}_g$ as the set of symmetric $g \times g$ matrices with positive definite imaginary parts. The symplectic group $Sp(g,\mathbb{Z})$ is the subgroup of $SL(2g,\mathbb{Z})$ such that all elements satisfy $M=J_g^t M J_g$ with $J_g$ being the canonical almost complex structure.

$Sp(g,\mathbb{Z})$ acts on $\mathbb{H}_g$ by $M(Z)=(AZ+B)(CZ+D)^{-1}$ where A,B,C and D are the block matrix entries of M.

Let $\rho : GL(g,\mathbb{C}) \to GL(V)$ be a rational representation on a finite dimensional $\mathbb{C}$-vector space then the associated modular forms are the holomorphic functions $f : \mathbb{H}_g \to V$ satisfying $f(M(Z))=\rho(CZ+D)f(Z)$ for all $M \in Sp(g,\mathbb{Z})$.

Cusps forms can be easily characterized as the elements of Siegel's $\Phi$ operator's kernel.

Modular forms in genus 2

If g equals 2 then the observed representations are the ones of $GL(2,\mathbb{C})$. We know from representation theory that all irreducible representations are isomorphic to a rep of the type $det^k \otimes Sym^{2j}(\rho_{standard})$.

We denote by $\rho_{standard}$ the standard representation $X \mapsto X$. $Sym^{2j}(\rho_{standard})$ is the associated symmetric product $GL(2,\mathbb{C}) \to Sym^{2j}(\mathbb{C}^2)$. $det$ is just the 1 dimensional determinant representation $GL(2,\mathbb{C}) \to \mathbb{C}$.

For $k\geq 4$ Tshushima has given a dimension formula for the vector space of cusps forms in An explicit dimension formula for the spaces of generalized automorphic forms with respect to Sp(2, Z). Proceedings of the Japan Academy, Ser. A, Mathematical Sciences, 59:139–142, 1983.

Satoh and Ibukiyama gave (but partly didn't publish AFAIK) generators for the modules of vector valued modular forms to the representations $det^k \otimes Sym^{2j}(\rho_{standard})$ with running k and fixed j in ${1,2,3}$.

The actual question

So the next question for me was are there cusps forms to $det^3 \otimes Sym^{2j}(\rho_{standard})$ and can they ( at least a single one) be given explicitly, in particular for j=4 ?

cheers Tom

p.s. please excuse all mistakes I made but it was the first time for me publishing on such a plattform.

Comment by Tom

2013-05-15T08:51:53Z

@ Peter. I was never claiming that I could solve the actual question. In fact, I stated that I can't. I said in the second line that I can give a linear polynomial (in an answer as I haven't got enough rep for commenting) and was then asked to elaborate which I did.

Comment by Tom

2013-05-14T20:07:42Z

I guess we are not on the same page. But, if I take $X$ to be a non-zero number -denoted by $x$- and $Y$ to be $x \cdot I_n$, then there is a solution, although $x$ has rank 1 and $Y$ rank n. Indeed, denoting by $e_i$ the i-th basis vector we have $$Y=\sum_i e_i \cdot x \cdot e_i^t=\sum_i x \cdot e_i \cdot e_i^t.$$ I hope haven't made new mistakes now.

Comment by Tom

2013-05-14T13:05:12Z

I have to admit that I am puzzled now. Peter, could you be so kind to give a short example ? Furthermore, wouldn't that make your comment the desired answer ? btw I just clarified the notation above.

Comment by Tom

2013-05-13T19:14:18Z

thanks for editing, Emil !

Comment by Tom

2013-03-11T18:03:21Z

Dear Aakumadula, thanks for helping with the syntax! Sorry, I haven't had time to look up your reference in detail, yet ! Dear Nathan, that's a really nice one ! But doesn't $u(1)$ generate all $u(x)$s ? And can't we pick only finitely many $C(a,b,c,d)$s as $\Gamma_0^{(1)}(N)$ is finitely generated ?

Comment by Tom

2013-03-05T18:05:38Z

Indeed, Mumford gives generators for $\Gamma_g$,$\Gamma_g[2]$ and $\Gamma_g[1,2]$ on pages 202-210. But to be honest I was hoping for more.

Comment by Tom

2013-03-05T17:22:31Z

Dear J, as you proposed I just clarified the notation $M=(A B \\ C D)$. With the few generators I was tkinking of E. Freitag 'Siegelsche Modulformen' (in Springer's Comprehensive Studies 254 ) appendix V pages 322-328. The proof relies on the fact that he finds for an EUCLIDEAN ring 'smaller' or 'easier to handle' sets of generators for $SL(g,R)$ and $GL(g,R)$. Now I'm having a look in the 2 books you mentioned.

Comment by Tom

2012-07-24T20:17:54Z

I just sent you an email.

Comment by Tom

2012-07-19T05:52:21Z

This is already a quite nice answer ! Do you know what kind of strategy he used ? Or do you know how to contact him or his supervisor ? To speak quite honestly I couldn't find anything on the Kyushu University homepage. I mean Satoh, Ibukiyama and their students collect Eisenstein series and sorts of Rankin Cohen brackets until they reach the dimension Tshushima has calculated 20 years ago. But this collection seems to me to be intricate especially if you raise j.