# Is there an explicit left invariant metric on the general linear group?

Consider $GL_n^+$, the group of (real) invertible matrices with positive determinant.

Is it possible to find an explicit formula for a metric on $GL_n^+$ which is left-invariant, i.e

$$d(A,B)=d(gA,gB) \, \,\forall A,B,g \in GL_n^+$$

and which generates the standard topology on $GL_n^+$.

(Without the last requirement the discrete metric will do).

Even finding a concrete example of a metric which is only "scale invariant" (satisfying $d(A,B)=d(rA,rB) \, \,\forall r \in \mathbb{R}$) will be an achievement. In fact even finding a metric which is invariant under multiplication by $r=2$ seems non-trivial.

A Riemannian approach:

One can take the metric $ds^2 = \text{tr}\bigl((g^{-1}dg)^Tg^{-1}dg\bigr)$ on $GL_n^+$ (this is the left translation of the standard Frobenius metric on $T_IGL_n^+ =M_n$), and to use its induced distance.

The question is if we can compute this distance explicitly.

For any symmetric positive-definite matrix $P$ $$d(I,P)=\|\log P\|_F \tag{*},$$ where $\|\cdot \|_F$ is the Frobenius norm, and $\log P$ is the unique symmetric logarithm of the matrix $P$.

This is proved here in section 3.3. The point is that it's easier to compute $d(A,\text{SO(n)})$ than $d(A,B)$; A minimizing geodesic from a point to a submanifold must intersect that submanifold orthogonally, hence we obtain more information which simplifies the analysis.

It can be shown that

$$d(A,\text{SO(n)})=d(A,Q(A))=\|\log \sqrt{A^TA}\|_F,$$ where $Q(A)$ is the orthogonal polar factor of $A$. In particular for positive matrices $P$, $Q(P)=I$, so we obtain $(*)$.

Calculating the distance $d(I,X)$ for an arbitrary is open for now.

Any such metric is determined by $f(X)=d(X,I)$, since $$d(A,B)=d(I,A^{-1}B)=f(A^{-1}B) \tag{1}$$

Transating the requirements from a metric we get that if $d$ is given in terms of $f$ as in $(1)$, then $d$ is metric if and only if

Positivity: $f(X)=0 \iff X=I \tag{2}$

Symmetry: $f(X)=f(X^{-1}) \tag{3}$

Triangle inequality: $f(XY) \le f(X) + f(Y) \tag{4}$

An equivalent formulation is thus following:

Find a non-negative function $f:GL_n^+ \to \mathbb{R}$ satisfying requirements $(2)-(4)$.

Reduction of the problem to $SL_n$:

Consider $f(X)=|\ln (\det X)|$. $\, \,f$ satisfies $(3),(4)$, and $f(X)=0 \iff X \in SL_n$.

Now, suppose we constructed a function $\tilde f:SL_n \to \mathbb{R}^+$ satisfying $(2)-(4)$ above.

Then, by defining $$\hat f(X)=f(X)+\tilde f(\frac{X}{\det(X)^{\frac{1}{n}}})$$

it is easy to see that $\hat f:GL_n^+ \to \mathbb{R}$ also satisfies $(2)-(4)$ as required.

Of course, an immediate way to "construct" a left-invariant metric is via left-invariant Riemannian metrics (which are very easy to build), but this usually** does not induce an explicit "formula-like" metric.

Also, is it true that in some sense the space of left-invariant metrics is "finite-dimensional"? (again I am talking on arbitrary metrics not just those which are induced by Riemannian metrics)

To make this notion more precise, some care should be taken. For instance, there are ways to generate new invariant metrics from old one (e.g. by $d \to \sqrt{d}$), but for this discussion we can identify two metrics if one is a function of the other.

I now think perhaps there is no way this space will be finite-dimensional. Since any left translation of a smooth norm, will induce a Finsler norm, and the space of smooth norms is not finite-dim in any reasonable way, we conclude that the space of metrics is also infinite-dim. (Since different Finsler norms give rise to different induced distances).

** For $n=1$, $GL_n^+=\mathbb{R}^{>0}$, and the fomula: $d(x,y)=|\ln(\frac{y}{x})|$ does the job. (It is in fact induced by the Riemannian metric obtained from left translation of the standrad metric on $T_1\mathbb{R}$). The obvious problem with generalizing this to higher dimensions is that there is no global matrix logarithm on $GL_n^+$, one has to choose a branch.

• I think this will do: $d(A,B)=\Big|\ln\big|\det\left(A^{-1}B\right)\big|\Big|$. I expect that this is the only metric, up to scalar multiple. – Batominovski Mar 19 '16 at 9:38
• Nice idea, but I think your metric is not positive. Take $A=Id$,$B$ diagonal with $2,\frac{1}{2}$ on the diagonal (or any $B \neq Id$ with determinant $\det B=1$). You will get $d(A,B)=0$ even though $A \neq B$. – Asaf Shachar Mar 19 '16 at 15:27
• Maybe my answer is the only $GL$-invariant "pseudometric" up to scalar multiple. Your question is very interesting, nonetheless. – Batominovski Mar 19 '16 at 20:02
• Is there a conformal left invariant Riemannian metric on general linear group?Is there another Lie group structure on $Gl(n,\mathbb{R}$ which is compatible to the standard smooth structure and admit a conformal Left invariant metric? – Ali Taghavi Aug 18 at 1:05

Let $P$ be a convex polytope in $R^n$ whose interior contains $0$ and which has no nontrivial linear symmetries (i.e. if $A$ is an invertible linear map, $AP=P$ implies that $A$ is the identity). In particular, $P\ne -P$. You can easily construct such $P$ by taking a suitable simplex or a cube. Then $P$ defines a nonsymmetric norm $||\cdot||$ on $R^n$ for which $P$ is the unit ball, by the usual procedure: $||v||=t$, where $t\in R_+$ is such that $t^{-1}v$ is on the boundary of $P$, and setting $||0||=0$. Using this norm we define the standard operator norm on linear endomorphisms of $R^n$: $$||A||= \max \{||Av||: v\in P\}.$$ This norm satisfies $||AB||\le ||A||\cdot ||B||$ and for every invertible matrix $A$, $$g(A)=\max(||A||, ||A^{-1}||)\ge 1$$ with equality if and only if $A=I$, the identity matrix. The function $g$ is "explicit" in the sense that $||A||$ is easily computable: It equals maximum of the norms $||Av_i||$, where $v_i$'s are the vertices of $P$. Now, set $f(A):= \log(g(A))$. This is your function. (All the required properties are clear.) If it is of any use, I do not know.
• Thanks. Your idea is very nice! A minor question: Why is $g(A)$ equal the maximum of the norm of $\|Av_i\|$? (It is clear to me why the maximum is obtained on the boundary of $P$, but not why it is always obtained at a vertex? (Is this is related to the fact $P$ has no symmetries? I did not see any other place where you used this assumption, but maybe I missed something). – Asaf Shachar Apr 27 '17 at 7:50
• @AsafShachar: It is a convex function, so its maximum on a convex set is attained at one of the extreme points of the subset (if you like, this a corollary of the Hahn-Banach theorem). The no-symmetry assumption is used to ensure that $f(A)=0$ implies that $A=I$. Think about what would happen if I were to take $P$ to be a ball centered at the origin. – Moishe Kohan Apr 27 '17 at 11:08
• Thanks!, I understand. Regarding convexity, how do you see that $g(A)$ is convex? It is not clear that the mapping $A \to \|A^{-1}\|$ is convex. (see here for instance). – Asaf Shachar Apr 27 '17 at 11:24
• @AsafShachar: I did not claim that the map $g: A\mapsto g(A)$ is convex (it is not). The claim is that the function $v\mapsto ||v||$ is convex, which is quite standard and easy. – Moishe Kohan Apr 27 '17 at 13:15