I was working on a homework problem that involved showing that the map $f:U(n)\rightarrow S^1,g\mapsto det(g)$ is a submersion (which is given here) And the following question emerged:
Given $g\in U(n)$ (where $U(n)$ is the group of unitary matrices), and $X\in T_I U(n)$, is it necessarily the case that $gX\in T_g U(n)$?
I believe this is true, indeed, since $X\in T_I U(n)$, there exists a smooth curve $\lambda:\mathbb{R}\rightarrow U(n)$ such that $\lambda(0)=I$ and $\lambda'(0)=X$. Define $\gamma(t):=g\lambda(t)$. Note that $\gamma$ is still a smooth map $\mathbb{R}\rightarrow U(n)$ since $U(n)$ is a group under multiplication.
Furthermore we do have $\gamma(0)=g\cdot \lambda(0)=gI=g$, and $$\frac{d}{dt}\gamma(t)=\frac{d}{dt}g\lambda(t) =g\frac{d}{dt}\lambda(t)$$ so it follows that $\gamma'(0)=g\lambda'(0)=gX$.
So we have shown that there exists a smooth curve $\gamma:\mathbb{R}\rightarrow U(n)$ such that $\gamma(0)=g$ and $\gamma'(0)=gX$ and so by definition $gX\in T_g U(n)$.
I know that this result seems straightforward but I am not yet comfortable working in this framework so I would appreciate if someone could just quickly glance through my work here and let me know if my reasoning is accurate. Thanks!
Here is the general idea. If $N$ is a submanifold of a manifold $M$ and a smooth map $f:M\rightarrow M$ satisfies the property that $f(N)\subseteq N$, then for all $x\in N$, $d_xf:T_xM\rightarrow T_{f(x)}M$ restricts to a linear map $T_xN\rightarrow T_{f(x)}N$. In your case, let us take $M=Mat_{n\times n}(\mathbb{C})$ and $N=U(n)$. Left-multiplication by a fixed $g\in U(n)$ gives $f=L_g$. Since $L_g$ is then a linear automorphism of $Mat_{n\times n}(\mathbb{C})$, $d_xL_g=L_g$ for all $x\in Mat_{n\times n}(\mathbb{C})$ if we identify tangent spaces to points in $Mat_{n\times n}(\mathbb{C})$ with $Mat_{n\times n}(\mathbb{C})$. In particular, $d_IL_g=L_g$ restricts to a linear map $T_{I}U(n)\rightarrow T_gU(n)$, so that $gX\in T_gU(n)$ for all $g\in U(n)$ and $X\in T_IU(n)$.