I was wondering if a for a matrix Lie group, an invariant metric can in general be thought of as the induced Metric of the group's submanifold in $\mathbb{R^n}^2$. This seems to line up for simple examples.
For example, in $U(1) = SO(2)$, the submanifold traces out a circle parameterized by the angle of rotation.* Also, in $SU(2)$, the submanifold can be thought of as the 3 sphere in $S^3 \subset \mathbb{R^4}$ if we think of $S^3$ as the group of unit quaternions, which gives us embedding in $\mathbb{R}^4$ that preserves the metric under the group action (i.e. multiplication by a unit quaternion as being a rotation in this 4d space).**
My question is does this notion of invariant measure as this induced metric carry over to all matrix Lie groups?
*,**: Clarification
I'm leaving out a detail, since for both the matrix representation of SO(2), $\theta \mapsto \begin{bmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \\ \end{bmatrix}$
and the (complex 2D) representation of SU(2) via quaternions $x_1 + x_2 i + x_3 j + x_4 k \mapsto \begin{bmatrix} x_1 + x_2 i & x_3 + x_4 i \\ -x_3 + x_4 i & x_1 - x_2 i \\ \end{bmatrix}$
The first column traces out the desired shape, which is a circle in $\mathbb{R^2}$ for $SO(2)$ and a 3-sphere in $\mathbb{R^4} = \mathbb{C^2}$ for SU(2). Then for both groups, the second column traces out the same shape in the second half of the coordinates. This means the metric on the group given by these maps is still the circle or 3-sphere metric multiplied by 2, roughly speaking. These are both still preserved by the group action.
EDIT****: I believe this doesn't work in general (see below comments) but does for unitary matrices. Note that the inner product of two matrices $A$ and $B$ is given by $d(A,B) = Tr(A^t B)$. So if we have a unitary matrix, U and consider the action L_U. Then $d(L_U (A), L_U (B)) = Tr((UA)^t(UB)) = Tr(A^t U^t U B) = Tr(A^t B) = d(A,B)$ since $U^t U = Id$. So the metric is invariant under the group action if the group is unitary.
Hint Consider a (say, left-)invariant metric $g$ on $\textrm{GL}(n, \Bbb R)$, which has the canonical natural embedding as an open subset into the space $M(n, \Bbb R) \cong \Bbb R^{n^2}$ of $n \times n$ matrices. Since $g$ is left-invariant, for all $A \in \textrm{GL}(n, \Bbb R)$, the left translation map $L_A : X \mapsto AX$ is an isometry of $g$, but we may identify the tangent map $T_I L_A$ at the identity $I$ with $A$ itself. Do the dilation maps, i.e., those corresponding to $A = \lambda I$, preserve the metric induced by the Euclidean metric on $\Bbb R^{n^2}$?