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In the coffee hour discussion today, Nick gave me an interesting explicit example of an exceptional isomorphism of Lie groups.

$G = \mathbb{SL}(4, \mathbb C) \to \mathbb{SO}(6, \mathbb C)$

is an isomorphism via the $\bigwedge^2$ map. Let me elaborate. The group $G$ acts naturally on $V = \mathbb C^4$. If $\{e_1, e_2, e_3, e_4\}$ is a basis of $V$, then a basis of $\bigwedge^2 V$ could be taken as $\{ e_1 \wedge e_2, e_1 \wedge e_3, \cdots, e_3 \wedge e_4 \}$. Thus, $\mathbb{GL}(\bigwedge^2 V)$ is $\mathbb {GL}(6, \mathbb C)$ and our $G$-action gives a monomorphism of groups:

$\bigwedge^2 : G=\mathbb{SL}(4,\mathbb C) \to \mathbb{GL}(\bigwedge^2 V) = \mathbb {GL}(6,\mathbb C)$.

The image must actually be inside $\mathbb{SL}(6,\mathbb C)$.

$(\; , \; ) : \bigwedge^2 V \times \bigwedge^2 V \to \mathbb C$,

$(e_i \wedge e_j , e_k \wedge e_l ) \mapsto e_i \wedge e_j \wedge e_k \wedge e_l$

is symmetric (two permutations) and preserves the $G$– action, i.e.,

$(g . \left(e_i \wedge e_j\right) , \left( g . e_k \wedge e_l \right) ) = (e_i \wedge e_j, e_k \wedge e_l)$.

Hence the image of $G$ in $\mathbb{GL}(6, \mathbb C) \simeq \mathbb{GL}(\bigwedge^2 V)$ must also preserve this symmetric bilinear form. Thus, $\bigwedge^2(G) \subseteq \mathbb{SO}(\bigwedge^2 V)$. By dimension consideration, they must be equal.

I was wondering the rationale behind naming parabolic subgroups of linear algebraic groups. The answer, interestingly, comes from the action of $SL(2,\mathbb R)$ on the upper half plane. (I came up with this little discovery on my own.)

The orbit of the point $i$ under the action of the standard parabolic subgroup of $SL(2,\mathbb R)$ is a parabola.

The upper half-plane is an object that comes up in many parts of mathematics, hyperbolic geometry, complex analysis and number theory to name a few. The group $SL(2,\mathbb R)$ acts on it by fractional linear transformations:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix} .\ z \mapsto \displaystyle \left( \frac{az+b}{cz+d} \right) .$

A parabolic subgroup is it’s subgroup $P$ such that quotienting by $P$ gives a compact variety. Upto conjugation, the only parabolic subgroup of $SL(2,\mathbb R)$ is $\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}.$ It’s action on the point $i$ is given by:

$\begin{pmatrix} x & y \\ 0 & x^{-1} \end{pmatrix} . i = \displaystyle \left( \frac{xi+y}{0i+x^{-1}} \right) = y + i x^2$

whose locus is a parabola.

I wonder why textbooks in algebraic groups don’t mention this!

EDIT: (21 April, 2015) The above calculation is WRONG. I don’t know the answer to “parabolic”.

I came across a simple statement in finite group theory that I’m almost upset no one told me earlier. The source is Serre’s book ‘Linear Representations of Finite Groups’. Serre used this statement below to define / prove Brauer’s theorem on induced representations of finite groups, using which one proves the meromorphicity of Artin-L-functions. Here it goes.

$G$ is a finite group and $p$ is a fixed prime. An element $x$ of $G$ is called $p$unipotent if $x$ has order a power of $p$ and $p$regular if it’s order is prime to $p$.

Cool result:  Every element $x$ in $G$ can be uniquely written as

$x = x_u x_r;$

where

• $x_u$ is $p$-unipotent, $x_r$ is $p$-regular,
• $x_u$ and $x_r$ commute and
• they are both powers of $x$.

The proof is really easy. Just replace $G$ by the (finite) cyclic group generated by $x$!

I call it the Jordan decomposition because we have a similar decomposition for endomorphisms (among other things).

Let $V$ be a finite dimensional vector space over an algebraically closed field of characteristic zero (just in case!). Each $x \in \text{End }(V)$ can be uniquely written as

$x = x_s x_u;$

where

• $x_s$ is semisimple (diagonalizable), $x_u$ is nilpotent,
• they both commute and
• they are polynomials in $x$ without a constant term.

Pretty cool, huh!

This post is just a collection of basic results I have compiled for referring to in desperate times Nothing too deep except Haar-von Neumann’s theorem.

A measure $\mu$ on a locally compact (Hausdorff, always) group $G$ is left-invariant if

$\boxed{ \int_G f(s^{-1}x) \text{d} \mu(x) = \int_G f(x) \text{d} \mu(x). }$

The most important theorem in this topic is the existence and uniqueness of the Haar measure (proven by Haar and von Neumann, respectively).

Theorem 1: On every locally compact group $G$ there is a unique (up to a positive constant of proportionality) left-invariant positive measure $\mu \neq 0$.

Proposition 2: $G, \mu$ as usual. For $G$ to have a finite measure, it is necessary and sufficient that $G$ be compact!

Proposition 3: There is a continuous group-homomorphism (called right modulus), $\triangle_r : G \to \mathbb R^*_+$ such that

$f(xs^{-1}) \text{d} \mu(x) = \triangle_r(s) \int f(x) \text{d} \mu(x).$

Proposition 4: Let $\mu$ (respectively, $\nu$) be a left-invariant Haar measure on locally compact groups $H$ (resp., $K$). Then the product integral on $G = H \times K$ is left-invariant and

$\boxed{\triangle^G_r(s,t) = \triangle^H_r(s) . \triangle^K_r(t).}$

Corollary 5: $G$ is unimodular (i.e., $\triangle_r =1$) precisely when $H$ and $K$ are so.

The most important example (for me) is the general linear group over number fields or $p$-adics and its subgroups. All semisimple (and more generally reductive) groups are unimodular. Compact groups are unimodular. Abelian groups are trivially so. However the measure on Borel (and parabolic) subgroups (i.e., upper triangular matrices) is not unimodular. The above proposition 4 allows one to transfer the Levi decomposition on the groups to their measures.

Let $G = GL_2(F)$ for some field $F$ and $B$ be the subgroup of upper triangular matrices in $G$. As usual, let $\mathbb P^1 = \mathbb P^1(F)$ denote the projective plane over $F$.

As $F$-varieties, $G/B$ is isomorphic to $\mathbb P^1$ because $B$ is the stabilizer of the usual (Mobius) action of $G$ on $\mathbb P^1$. The resulting action of $G/B$ on $\mathbb P^1$ is 1-transitive and fixed-point free. Indeed, counting points when $F$ is a finite field with $q$ elements agrees with the facts that

$\# G(\mathbb F_q) = (q^2-1)(q^2-q)$

and

$\# \mathbb P^1(\mathbb F_q) = q+1$.

This generalizes in that parabolic subgroups (closed subgroup containing a Borel subgroup, i.e., containing the upper triangular matrices upto conjugation, for the general linear group) of an algebraic group give rise to projective varieties. They are precisely the stabilizers of a flag and conversely, the set of flags can be given the structure of an $F$-space and in particular a topology wherein they are compact spaces (being projective spaces). They are called flag varieties.

In my recent Riemann surfaces class, Donu started talking about de Rham cohomology and it’s generalization to Hodge theory. I was fascinated by it (in particular, the Hodge decomposition) so started reading some related stuff. I initially intended to write a blog post about it, but soon it grew in size than I had intended, so I’m attaching it to this post as a pdf file. Talking to Nick and Partha and looking at Forster’s Lectures on Riemann Surfaces was also helpful.

A mild introduction to Hodge_theory.

In this post, I will give a counter-example, that the image of a hyperspecial maximal compact subgroup under a surjective map need not (even) be maximal. In particular, this will tell us that a maximal compact subgroup (e.g., $SO(n)$ inside $SL_n(\mathbb R)$) need not be maximal under surjection.

Let $G$ be a connected reductive algebraic group over a local non-archimedian field $F$. A subgroup $K$ of $G(F)$ is called hyperspecial maximal compact subgroup if

• $K$ is a maximal compact subgroup of $G(F)$,
• There is a group scheme $\mathcal G$ such that $\mathcal G(\mathcal O_F)=K$ and $\mathcal G(\mathcal O_F / \varpi \mathcal O_F)$ is a connected reductive group over $\mathbb F_q$, where $\varpi$ is a uniformizer in $\mathcal O_F$ and $q$ is the cardinality of the residue field.

Denote by $\phi$ the canonical surjection

$\phi : SL_2 \to PGL_2 = GL_2 / G_m$

(over some local field $F$, or surjection of group schemes, to be pedantic). Choose a non-unit element $x$ of $\mathcal O_F$. Then

$\begin{pmatrix}x & \\ & x^{-1} \end{pmatrix} \not\in SL_2(\mathcal O_F)$,

but this matrix is in $PGL_2(\mathcal O_F)$ because

$\begin{pmatrix} x & \\ &x^{-1} \end{pmatrix} = \begin{pmatrix} x & \\ & x^{-1} \end{pmatrix} . \begin{pmatrix} x & \\ & x \end{pmatrix} = \begin{pmatrix} x^2 & \\ & 1 \end{pmatrix}$.

Hence the image $\phi(SL_2(O_F) \subsetneq PGL_2(O_F)$ is compact but not maximal compact (so in particular, not hyperspecial).

Remarks:

• A good reference to read about hyperspecial maximal compact subgroups is Tits’ article in Corvallis (Volume 1). Nick and I are planning to start a reading seminar next semester in Spring 14 on the 40-odd page article.
• I found this counter-example somewhere on MathOverflow, while I was preparing the Borel-Tate articles (Corvallis, Volume 2) to speak in our “Arthur seminar”.

In my recent number theory seminar on “Hilbert’s 90 and generalizations” (notes here), Professor Goins asked the following interesting question.

Let ${K}$ be a field and ${d\in K^*}$. Define ${T_d}$ to be the torus

$\displaystyle \left\{ \begin{pmatrix} x & dy \\ y & x \end{pmatrix} : x,y \in L, x^2-dy^2=1\right\}.$

What values of ${d}$ give ${K}$-isomorphic tori?

(The question was perhaps motivated by the observation that over the reals, the sign of ${d}$ determines completely whether ${T_d}$ would be split (i.e., isomorphic to ${\mathbb R^*}$) or anisotropic (i.e., isomorphic to ${S^1}$).

Here are two ways of looking at the answer.

• For ${d,e \in K^*}$, we determine when two matrices ${\displaystyle\begin{pmatrix} x & dy \\ y & x \end{pmatrix}}$ and ${\displaystyle\begin{pmatrix} u & ev \\ v & u \end{pmatrix}}$ are conjugate in ${\text{SL}_2(K)}$. Solving the system

$\displaystyle \begin{pmatrix} x & dy \\ y & x \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} . \begin{pmatrix} u & ev \\ v & u \end{pmatrix} . \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

gives ${\displaystyle \frac{e}{d} = \left(\frac{b}{c}\right)^2, de = \left(\frac{d}{a}\right)^2}$.

Thus ${e \in d.(K^*)^2}$ i.e., the ${T_d}$‘s are classified by ${\displaystyle \frac{K^*}{(K^*)^2}}$. (For ${K=\mathbb R}$, this is isomorphic to ${\{\pm 1\}}$ so the sign of ${d}$ determines ${T_d}$ upto conjugation.) By Kummer theory, ${\displaystyle \frac{K^*}{(K^*)^2} \cong H^1(\text{Gal}(\overline K / K), \mu_2)}$, where ${\mu_2 = \{\pm 1\}}$ are the second roots of unity. Thus there is a correspondence between isomorphism classes of tori ${T_d \; (d \in K^*)}$ and quadratic extensions of ${K}$.

• Another way to look at the same thing is as follows. Fix ${d \in K^*}$. Let ${L}$ be an extension of ${K}$ wherein ${T_d}$ splits. Now ${T_d(L)}$ is a split torus of rank 1. For an algebraic group ${G}$ over an algebraically closed field, we have the exact sequence

$\displaystyle 1 \rightarrow \text{Inn}(G) \rightarrow \text{Aut}(G) \rightarrow \text{Aut}(\Psi_0(G)) \rightarrow 1,$

where ${\Psi_0(G)}$ is the based root datum ${(X,\Delta,X\;\check{}, \Delta\,\check{})}$ associated to ${G}$. (Here, ${X = X^*(G)}$ and ${\Delta}$ is the set of simple roots of ${X}$ corresponding to a choice of a Borel subgroup of ${G}$.) For details, see Corollary 2.14 of Springer’s paper “Reductive Groups” in Corvallis.

In our case, ${G = T_d}$ so ${\Psi_0(G) = ( \mathbb Z, \emptyset, \mathbb Z \;\check{}, \emptyset)}$.

$\displaystyle \text{Aut}(\Psi_0(G)) \cong \text{Aut}(\mathbb Z) \cong \{ \pm 1\}.$

Now ${L/K}$ forms of ${T_d}$ are in bijective correspondence with

$\displaystyle H^1(\text{Gal}(L/K), \text{Aut}(\Psi_0(G))) = H^1(\text{Gal}(L/K), \{\pm 1\}) \cong \text{Hom}_{\mathbb Z}(\text{Gal}(L/K), \{\pm 1\});$

the last isomorphism because the Galois group acts trivially on the split torus ${T_d(L)}$. ${\blacksquare}$

In this previous post, we saw the existence of a common eigenvector, namely $\phi(n) = a_n =$ number of nonzero solutions to $x^2=d$ modulo $n$. This was not a coincidence. Indeed, it was based on the fact that $\{ T_p : p \nmid N := 4|d| \}$ is a family of self-adjoint and commuting operators on the space of complex-valued functions on $G = (\mathbb Z/N \mathbb Z)^*$.

$\langle f,g \rangle = \displaystyle\sum_{r\in G} f(r) \overline{g(r)} . \qquad )$

This post is a generalization.

Let $f$ be a holomorphic function on the upper half complex plane. We say $f$ is modular if it satisfies a technical condition called “holomorphic at the cusps” and the following.

$\displaystyle f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k f(z) \quad \forall \gamma = \begin{pmatrix} a&b\\c&d \end{pmatrix} \in \Gamma := SL(2,\mathbb Z).$

Given any $f$ holomorphic on the upper half plane and $\gamma \in \Gamma(1)$, define

$\displaystyle (f|_\gamma)(z) := (cz+d)^{-k} (\text{det} \gamma)^{k/2} f\left(\frac{az+b}{cz+d}\right), \quad \forall \gamma = \begin{pmatrix} a&b\\c&d \end{pmatrix} \in \Gamma(1)$.

It is a fact that for any $\alpha \in \text{GL}(2,\mathbb Q)^+$, there is a double-coset decomposition $\Gamma(1) \alpha \Gamma(1) = \displaystyle\bigcup_{i=1}^l \Gamma(1) \alpha_i$.

Define for such a decomposition,

$f|_{T_\alpha} := \displaystyle\sum_{i=1}^l f|_{\alpha_i}$.

Observe that $(f|_{\alpha})|_{\beta} = f|_{\alpha\beta}$ so that defines a well-defined action of $\Gamma(1)$ on the $f$‘s. There is a vector space called the space of modular forms and a $T_\alpha$-invariant subspace – $S_k(\Gamma(1))$ – the space of cusp forms (similar to $V_N$ in the previous post) and for varying $\alpha$, the operators $T_\alpha$ (called the Hecke operators)

$T_\alpha : S_k(\Gamma(1)) \to S_k(\Gamma(1))$,

$T_\alpha(f) := f|_{T_\alpha}$.

It’s a cool theorem that the Hecke algebra is commutative and the Hecke operators are self-adjoint with respect to an inner product (the Petersson inner product). A standard result in linear algebra tells that these can be diagonalized; there is a common eigenvector, called the Hecke eigenform. When suitably normalized, it’s associated $L-$ function has an Euler product (similar to the $\zeta$ function). This Euler product gives the Ramanujan’s identity –

$\displaystyle\sum_{n=1}^\infty \tau(n) n^{-s} = \prod_p \frac{1}{1-\tau(p) p^{-s} + p^{11-2s}}.$

(Here, $\tau$ is the Ramanujan-$\tau$ function. )

Pretty cool stuff, eh!

——————————————————————————————————————

Tailpiece: References (since I am very vague here) –

• A first course in modular forms – Diamond, Shurman
• Automorphic forms and representations – Daniel Bump

Also, I was interested in the properties the L-function corresponding to the $a_p$‘s in the earlier post. I haven’t seen any book that mentions about these.

Below is Gauss’ quadratic reciprocity, found in most elementary texts in number theory. In this post, we’ll see how the Hecke operators originate from this theorem.

Theorem (Gauss). Let $\varepsilon(n) = (-1)^{\frac{n-1}{2}}$ and $\omega(n) = (-1)^{\frac{n^2-1}{8}}$. For distinct odd primes $p, q$,

$\displaystyle \left(\frac{p}{q}\right) = \varepsilon(p) .\varepsilon(q). \left(\frac{q}{p}\right),$

$\displaystyle \left(\frac{-1}{p}\right) = \varepsilon(p),$

$\displaystyle \left(\frac{2}{p}\right) = \omega(p).$

Consider the equation

$\textbf{(Q)}: \qquad x^2 = d; \qquad \qquad d\in \mathbb Z \backslash \{0\}.$

Let $a_p(Q)$ be the number of solutions to Q modulo $p$, minus one. Then by definition of the Legendre symbol, $a_p(Q) = \left( \frac{d}{p} \right).$ By property of the Lengendre symbol (or rather, the Jacobi symbol), we have

$a_{mn}(Q) = a_m(Q) . a_n(Q).$                       (*)

Let $N = 4 |d|$. Then it follows from the reciprocity law that $a_p(Q)$ depends only on the value of $p$ modulo $N$. Furthermore, the finite sequence ${a_2(Q), a_3(Q), a_5(Q), \cdots }$ arises as a set of eigenvalues of a linear operator (the Hecke operator) on a finite dimensional complex vector space. We’re going to construct the space.

$V_N := \{ f: (\mathbb Z/n\mathbb Z)^* \to \mathbb C \}$

$T_p : V_N \to V_N, \quad T_p(f)(n) = f(pn) \quad \text{if } p \nmid n \text{ and } 0 \text{ otherwise }.$

Verify that $T_p$ is a linear operator on $V_N$. Now all these operators for varying primes commute with each other. So what is a common eigenvector?

Define $\phi (n) = a_n(Q)$. Then use (*) to show that

$T_p(\phi) = a_p(Q) \phi,$

for all $p$ prime. So, this $\phi$ is indeed, a common eigenvector for all the $T_p$‘s!

I will explain more about Hecke operators on modular forms in a future post. (Edit July 10, 2013: Link to the said post).

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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