# Constructing a metric on a Lie Group

Given a connected Lie Group $G$ is there a standard method for constructing the metric tensor?

One can of course expand a group element $g=e^{x^ie_i}$ where $e_i$ are basis vectors of the Lie algebra about the identity up to linear order and compute the metric in this manner, but is there a more direct way?

An example would also be appreciated.

There is such a (bi-invariant) metric on a Lie group (isomorphic to a compact Lie group times $$\mathbb{R}^{n}$$). It is the one induced by the Killing form. For more input on this, see the mathoverflow discussion on the topic:

https://mathoverflow.net/questions/32554/why-the-killing-form

It should be pointed out that you do not need the exponential map to construct the metric, which is defined on the tangent space $$T_{p}g, g\in G$$, not on the group itself. All you need is a nicely behaved symmetric bilinear form under the group action. The explicit construction can be found at here.

Update: Yor pointed out in the comments that for a non-semisimple compact Lie group there is a bi-invariant metric, but it is not induced by the Killing form, as the Killing form degenerates on the center. To solve this, write $$\mathfrak{g}=\mathfrak{a}\times \mathfrak{s}$$. The Killing form induces a metric on $$\mathfrak{s}$$ and we put an arbitrary metric on $$\mathfrak{a}$$.

• I don't think Lie groups in general admits a bi-invariant metric though.
– Qidi
Dec 17, 2016 at 23:22
• @Qidi: Yes, compactness is needed. Dec 17, 2016 at 23:24
• @Bombyxmori: Compactness is useful, but not required: all abelian groups admit bi-invariant metrics! [A Lie group that admits a bi-invariant metric is the product of a compact group and a copy of some $\mathbb{R}^m$ (Milnor, 1976).] Dec 18, 2016 at 0:56
• @CatalinZara: I only know the proof in the compact case. If I am not confused, Milnor's claim is that a lie group admits a bi-invariant metric can be decomposed as a product group, not every (connected) Lie group admits a bi-invariant metric. I have yet to read the paper itself, so thanks for the pointer. Dec 18, 2016 at 1:32
• @Bombyxmori: What I'm saying is that there are groups that admit bi-invariant metrics and are not compact. Dec 18, 2016 at 18:12

You ask about the metric tensor of a Lie group, but such a thing does not exist. What you can do, though, is to (arbitrarily) choose a scalar product $$\langle, \rangle _e$$ in $$T_e G$$ and then by (left) translations define $$\langle, \rangle _g = (L_g) _* \langle, \rangle _e$$, that is for $$u,v \in T_g G$$ define $$\langle u, v \rangle _g = \langle \mathrm d L_g ^{-1} (u), \mathrm d L_g ^{-1} (v) \rangle _e$$, where $$L_g (x) = gx$$. This is a left-invariant Riemann tensor, which in turn gives rise to a left-invariant distance $$d$$, i.e. $$d(gx, gy) = d(g,h)$$ for all $$g,x,y \in G$$. Neither the metric tensor, nor the associated distance need be right-invariant (and, symetrically, if you construct them to be right-invariant, they need not be left-invariant).

The following result is taken from Chapter 17 of "Linear Algebra, Manifolds, Riemannian Metrics, Lie Groups and Lie algebra, with Applications to Robotics, Vision and Machine Learning" by Jean Gallier:

If $$G$$ is connected, $$\langle, \rangle$$ defined above is bi-invariant (meaning left- and right-invariant) if and only if $$\langle [u,v], w \rangle = - \langle v, [u,w] \rangle$$ for all $$u,v,w \in T_e G$$.

If $$G$$ is compact, then one may define a bi-invariant Riemann tensor $$\langle, \rangle '$$ from a left- (or right-) invariant one $$\langle, \rangle$$ by Haar-integrating the latter: $$\langle u, v \rangle ' = \int _G \langle \mathrm d L_g u, \mathrm d L_g v \rangle_{gx} \ \mathrm d g$$ for all $$u,v \in T_x G$$ and $$x \in G$$.

If $$G$$ is semisimple, then its Lie algebra $$\mathfrak g \simeq T_e G$$ possesses a preferred scalar product - the Kiling form (or its opposite, depending on the sign convention used), and this in turn gives rise to a preferred metric tensor on $$G$$.

• Excuse me... would you please give me a help? So... it seems like you constructed a metric by choosing an inner product on $T_e G$, then passing out to other points $g\in G$ via left translation $L_g$, right? Note that a metric is a smooth positive covariant 2-tensor. So... my question is why the metric $g\to L_g$ is smooth?
– Eric
Nov 12, 2021 at 15:42