# How do the properties of a Lie group (represented as a manifold) manifest in the metric tensor of that manifold?

In general there is no reason why a metric $$g$$ on a Lie group $$G$$ to manifest any of the group properties. For example, you might know from your study of differential geometry that every smooth manifold $$M$$ can be endowed with a Riemannian metric $$g$$ (essentially just by pulling back the metric on $$\mathbb R^n$$ via charts and gluing them together). There is no reason whatsoever that this generic construction to obey any properties of group multiplication if you choose $$M=G$$.

However, if the Lie group is nice enough (e.g. compact) you can have metrics, which are for example left- (or even bi-) invariant. For example, what you can do is the following. Assume that you choose your favourite scalar product $$\left< \cdot, \cdot \right>$$ on $$\mathbb R^n$$. Then since $$T_e G \cong \mathbb R^n$$, we have a scalar product on the tangent space at the identity and then define for all elements $$x \in G$$

$$\forall v,w \in T_x G \quad (v,w)_x := \left$$

where $$L_x$$ is multiplication on the left by $$x \in G$$ and $$DL_x$$ is the differential of $$L_x$$. So what you do is pull any vector $$v \in T_x G$$ by the left multiplication back to $$T_e G$$, compute the scalar product there. Note that this scalar product will generally not be bi-invariant. If $$G$$ is compact, then you can "average" this scalar product with the (left invariant) Haar measure to get a bi-invariant scalar product. In fact, bi-invariant scalar products do speak volumes about the group structure of $$G$$.

I think I understand what you are asking. Assume you have a Lie group $$G$$ of dimension $$n$$ with a left, or right, or bi invariant metric $$g_0$$ on $$G$$. This means that either $$(L_q)^*g_0 = g_0$$ or $$(R_q)^*g_0 = g_0$$ or both for all $$q \in G$$.

1. First, you take a (possibly small) open subset $$U$$ of $$G$$.
2. Then, you take a diffeomorphism $$\phi : U \to W \subset \mathbb{R}^n$$ that maps the open domain $$U$$ of $$G$$ to an open subset $$W$$ of $$\mathbb{R}^n$$ with coordinates $$x = (x^1, ..., x^n)$$.
3. And, finally, you send the metric $$g_0$$ via $$\phi$$ from the Lie group $$G$$ to the open set $$W$$ in $$\mathbb{R}^n$$. More precisely, you define the (pull back) metric tensor $$g(x) = \big(\phi^{-1}\big)_x^*g_0(\phi^{-1}(x))$$ on the open set $$W$$ in $$\mathbb{R}^n$$ with coordinates $$x = (x^1, ..., x^n)$$.

Basically, $$\big( U, \phi\big)$$ is a chart of $$G$$ and $$g(x)$$ is the coordinate representation of $$g_0$$ in that chart.

Now, your question seems to be about the converse. Imagine you give me just a metric tensor $$g(x) = \big(\, g_{ij}(x)\,\big)_{i,j = 1}^{n}$$ defined on some open domain $$W$$ of $$\mathbb{R}^n$$ and you ask me: 'Can you figure out whether the metric tensor $$g(x)$$ is the coordinate representation of a left, or right or bi-invariant metric on some Lie group, i.e. did the metric tensor $$g(x)$$ came from a Lie group after the execution of procedure 1-2-3 from above?'

To answer this question, what I would do is to try and find the Lie algebra of Killing vector fields of $$g(x)$$, i.e. I will look for all vector fields $$X(x)$$ for which the Lie derivative of $$g(x)$$ is $$L_Xg = 0$$. Equivalently, if in coordinates $$x = (x^2,...,x^n)$$ on $$\mathbb{R}^n$$, the vector fields $$X(x)$$ are written as $$X(x) = X^{j}(x^1,...,x^n)\, \frac{\partial}{\partial x^j}\,$$ then the equation $$L_Xg = 0$$ can be written as the system of equations $$\nabla_{\frac{\partial}{\partial x_k}}\,X_j(x) \, + \,\nabla_{\frac{\partial}{\partial x_j}}\,X_k(x) = 0$$ So I would try to form the set of all such Killing vector fields $$\mathcal{K} = \big\{\, X(x) \, : \, L_Xg = 0 \, \big\}$$ This set $$\mathcal{K}$$ is in fact a finite dimensional Lie algebra of vector fields that generate the isometries of the metric tensor $$g(x)$$. If $$\dim{\mathcal{K}} < n$$ then there is no chance that $$g(x)$$ comes from a metric $$g_0$$ on a Lie group $$G$$. However, if $$\dim{\mathcal{K}} = n$$ then it does. Indeed, according to Cartan-Lie theorem (maybe also called Lie's third theorem), every finite dimensional Lie algebra is the Lie algebra of a simply connected finite dimensional Lie group. The appropriate combination of phase flows of a basis of Killing vector fields from $$\mathcal{K}$$ will generate the diffeomorphism $$\phi$$ discussed in point 2 above.

Finally, if I am not wrong, I believe that if $$\dim{\mathcal{K}} > n$$ it's probably because the metric tensor $$g(x)$$ comes from the metric of a homogeneous space. For example, the two-sphere is a homogeneous space for $$SU(2)$$, i.e. $$SU(2)$$ acts nicely on the two sphere and the two-sphere is diffeomorphic to the quotient of $$SU(2)$$ by a copy of U(1) embedded in $$SU(2)$$. For example, recall the Hopf fibration map. That is exactly this quotient map from $$SU(2)$$ to the two-sphere.