Is there a bi-invariant metric (not Riemannian) on $GL_n$?

Question

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 $\det >0$).

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

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

Answer 1

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

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.

Answer 2

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 $K$.

Answer 3

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 $m > 2$.