1
$\begingroup$

Is there a bi-invariant* metric $d$ on $GL_n$ (the group of invertible marices) which generates the standard topology on it? (the subspace topology from $\mathbb{R}^{n^2}$)

Note: I mean any metric (in the sense of metric spaces), not just metrics which come form Riemannain metrics. (It's a known result that Riemannian bi invariant metrics do not exist on $GL_n^+$).

(I am not interested in problems which might arise from connectedness issues, so if it becomes relevant, think instead on $GL_n^+$ = invertible with $\det >0$).


*bi-invariant metric is one which is invariant under left and right translations:

$d(A,B)=d(gA,gB)=d(Ag,Bg) \, \forall A,B,g \in GL_n$

$\endgroup$
1
$\begingroup$

No. A connected Lie group admits a bi-invariant metric iff it's isomorphic to $K\times \mathbb{R}^n$ for some compact Lie group $K$.

$\endgroup$
  • $\begingroup$ Thanks. But I meant to allow arbitrary metrics (in the sense of metric spaces), not just Riemannian. I wil edit the question accordingly. $\endgroup$ – Asaf Shachar Oct 29 '15 at 20:17
  • $\begingroup$ By the way, do you see an immediate way to see why $GL_n^+$ cannot be isomorphic to such a product? $\endgroup$ – Asaf Shachar Oct 29 '15 at 20:20
  • 1
    $\begingroup$ Well, there's always the metric $d(g, h)= \delta_{gh}$, which is trivially $G$-invariant. $\endgroup$ – anomaly Oct 29 '15 at 20:34
  • $\begingroup$ Consider the centralizer of an arbitrary element of $G$. $\endgroup$ – anomaly Oct 29 '15 at 20:36
  • 1
    $\begingroup$ No, I meant to allow any metric in the sense of metric spaces, and the trivial metric is $G$-invariant, however , it does not generate the standard topology on $GL_n$. $\endgroup$ – Asaf Shachar Oct 29 '15 at 20:41
1
$\begingroup$

$\newcommand{\lam}{\lambda}$ $\newcommand{\R}{\mathbb{R}}$

Credit: I heard this solution from Amotz Oppenheim.

Assume by contradiction there is such a metric $d$. Consider the following matrices: $D=\begin{pmatrix} \lam_1 & 0 & \cdots & 0 \\ 0 & \lam_2 & \cdots & 0 \\ \vdots & \vdots& \ddots & \vdots \\ 0 & 0 & \cdots & \lam_n \end{pmatrix} , A=\begin{pmatrix} 1 & a_1 & 0 & \cdots & 0 \\ 0 & 1 & a_2& \cdots & 0 \\ 0 & 0 & 1& \ddots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots & \\ 0 & 0 & 0& \cdots & 1 \end{pmatrix}$ $D$ is diagonal, $A$ is a matrix with all entries zero, except on the diagonal and directly above it. Since left (right) multiplication by a diagonal matrix amounts to multiplying each row (column) by the corresponding diagonal element we obtain:

$DA=\begin{pmatrix} \lam_1 & \lam_1a_1 & 0 & \cdots & 0 \\ 0 & \lam_2 & \lam_2a_2& \cdots & 0 \\ 0 & 0 & \lam_3& \ddots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots & \\ 0 & 0 & 0& \cdots & \lam_n \end{pmatrix} \Rightarrow DAD^{-1}=\begin{pmatrix} 1 & \frac{\lam_1}{\lam_2}a_1 & 0 & \cdots & 0 \\ 0 & 1 & \frac{\lam_2}{\lam_3}a_2& \cdots & 0 \\ 0 & 0 & 1& \ddots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots & \\ 0 & 0 & 0& \cdots & 1 \end{pmatrix}$

So $DAD^{-1}$ has $1$'s on the diagonal, and above it: $(DAD^{-1})_{i,{i+1}}=\frac{\lam_i}{\lam_{i+1}}a_i:=b_i$ Hence $D^2AD^{-2}$ has also $1$'s on the diagonal and $(D^2AD^{-2})_{i,{i+1}}=\frac{\lam_i}{\lam_{i+1}}b_i=(\frac{\lam_i}{\lam_{i+1}})^2a_i$ above it. Similarly $(D^nAD^{-n})_{i,{i+1}}=(\frac{\lam_i}{\lam_{i+1}})^na_i$

In other words, conjugation leaves the diagonal invariant and multiplies each entry above the diagonal by the ratio of the corresponding diagonal elements of the conjugate matrix.

If we choose for instance all $a_i=1$ and $\lam_i$ such that $\frac{\lam_i}{\lam_{i+1}}=\frac{1}{2}$, then $D^nAD^{-n}$ converges entrywise to the identity $I$, hence converges w.r.t the subspace topology on $GL_n$ induced by $\R^{n^2}$.

Since we assumed $d$ generates the topology on $GL_n$ it follows that $D^nAD^{-n}$ converges to $I$ also w.r.t $d$.

Now the assumed bi-invariance implies:

$d(A,I)=d(DAD^{-1},I)=d(D^nAD^{-n},I) \to 0$

So $D(A,I)=0$ which is a contradiction.

$\endgroup$
0
$\begingroup$

If you're interested in an arbitrary invariant topological (rather than Riemannian) metric $d$ that reproduces the usual topology, put \begin{align*} g_n &= \begin{pmatrix} n & 1\\ 0 & 1\end{pmatrix} & h_n &= \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \end{align*} Then $d(g_n h_n, 1) = d(g_n, h_n^{-1}) = d(h_n g_n, 1)$ for all $n$. Since $d$ is equivalent to the usual metric on $GL_2(\mathbb{R})$, we thus have \begin{align*} \|g_n h_n - 1\| \leq C \|h_n g_n - 1\|, \end{align*} for some constant $C > 0$, where $|\cdot|$ denotes the usual norm on $GL_2(\mathbb{R})$. But \begin{align*} g_n h_n &= \begin{pmatrix} n & n+1 \\ 0 & 1\end{pmatrix} & h_n g_n &= \begin{pmatrix} n & 2 \\ 0 & 1\end{pmatrix}, \end{align*} and taking $n$ sufficiently large gives a contradiction. The result therefore also holds for $GL_m(\mathbb{R})$ with $m > 2$.

$\endgroup$
  • 1
    $\begingroup$ I think $(h_ng_n)_{11} = n$... $\endgroup$ – Asaf Shachar Oct 30 '15 at 17:17
  • $\begingroup$ Thanks, you're right. I took another stab at the computation. $\endgroup$ – anomaly Oct 30 '15 at 20:07
  • $\begingroup$ Sorry, but that does not work either, since: $\| g_nh_n - Id\|^2=(n-1)^2 +(n-1)^2$ while $\| h_ng_n - Id\|^2=(n-1)^2 +4$, hence it's not hard to verify (by direct computation) that $\|g_n h_n - Id\| \leq C \|h_n g_n - Id\|$ forc $C=\sqrt 3$. $\endgroup$ – Asaf Shachar Oct 31 '15 at 9:22
  • $\begingroup$ I am also not sure how you deduced the existence of such a $C$ from the assumption that the invariant metric generates the same topology on $GL$. This does not necessarily imply strong equivalence between the two metrics. $\endgroup$ – Asaf Shachar Oct 31 '15 at 9:26
  • $\begingroup$ If you are interested, I have published a solution (see my answer above), which uses certain choices of conjugation, essentially like you suggested. $\endgroup$ – Asaf Shachar Dec 19 '15 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.