I have proved that $U(n)$, which is a group of unitary matrices, is a smooth manifold of real Dimension $n^2$. Now I am trying to show that $SU(n)$, which is a group of unitary matrices with determinant 1, is a smooth submanifold of $U(n)$ of real dimension $n^2-1$. However I am stuck at a step:
The idea is to use the theorem: Suppose $f:N^n \to M^m$ is a smooth map between smooth manifolds $N,M$ of dimension $n,m$ respectively. Then preimage of any regular value $y$ of $f$ is a regular submanifold of $M$ and is of dimension $n-m$. Here, $y$ is a regular value means that the rank of $f$ at every point in the preimage of $y$ has rank exactly $m$.
To use this Theorem, I considered the determinant map
$$ det: U(n) \to S^1:=\{z \in \mathbb{C}:|z|=1\}. $$ Now I want to show that
$1 \in S^1$ is a regular value of the map $det.$ That is, I have to show that derivative of $det$ at an arbitrary $g\in det^{-1}(1)=SU(n)$ has rank $1$. For an arbitrary $Y \in Mat_{n \times n}\mathbb{C}$, I get $$D_gdet(Y)=\lim_{t \to 0}\frac{det(g+tY)-det(g)}{t}=\lim_{t \to 0}\frac{det(g+tY)-1}{t}$$
I don`t know how should I proceed from here. How can I simplify the RHS?
Best Answer
You need to calculate the derivative of $\det$, which is
$$D_g \det (Y) = \det (g) \text{tr}(g^{-1}Y) = \text{tr} (g^{-1} Y) \in i\mathbb R = T_1\mathbb S^1 $$
as $\det (g) = 1$. Note that it suffices to find, for all $g\in \det^{-1}(1)$, $Y\in T_gU(n)$ so that $D_g\det(Y) \neq 0$. Note that
$$T_gU(n) = g\cdot u(n),$$
where $u(n)$ is the Lie algebra of $U(n)$ given by
$$u(n) = \{ A \in M_n(\mathbb C) : A + A^* = 0\}$$
Then $iI_n \in u(n)$ and $Y = g(iI_n)$ satisfies
$$D_g \det (Y) = \text{tr} (g^{-1} g(iI_n)) = ni\neq 0.$$