# 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 $x \in G$

$$\forall v,w \in T_x G \quad (v,w)_x := \left<DL_x^{-1} v,DL_x^{-1} w\right> $$

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 $q \in G$.

- First, you take a (possibly small) open subset $U$ of $G$.
- 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 $x = (x^1, ..., x^n)$.
- 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.