$SL(n,\Bbb{C})$ is a regular submanifold of $GL(n,\Bbb{C})$

Question

Let $SL(n,\Bbb{C})$ be the group of matrices of complex entries and determinant $1$.

I want to prove that $SL(n,\Bbb{C})$ is a regular submanifold of $GL(n,\Bbb{C})$.

An idea is to use the Regular Level Set Theorem because $$SL(n,\Bbb{C})=f^{-1}(1)$$ where $f(A)=\det(A)$

(I know how to prove the same for $SL(n,\Bbb{R})$ but here i believe we need to use theory of multivariable complex analysis which i do not know at all.)

Can someone help me prove this statement?

Thank you in advance.

Answer 1

You can avoid the calculations by using Lie group theory, because $GL(n,\mathbb{C})$ is a Lie group and $SL(n,\mathbb{C})$ a closed subgroup. Cartan's theorem then says $SL(n,\mathbb{C})$ is a regular Lie subgroup, in particular a regular submanifold. Note that this also works for $\mathbb{R}$ instead of $\mathbb{C}$.

Answer 2

You have to prove that $1$ is a regular value of $\det$. If $A(t)$ is a curve of matrices in ${\rm GL}(n,\Bbb C)$, then we have the formula

$$\frac{{\rm d}}{{\rm d}t} \det A(t) = \det A(t) \,{\rm tr}(A(t)^{-1}\dot{A}(t)).$$So we have to show that if $A \in {\rm SL}(n,\Bbb C)$, then ${\rm d}(\det)_A$ is onto. As it is scalar-valued and linear, it suffices to show that it is not the zero functional. For this reason, assume that $A(0) = A$ and $\dot{A}(0) = \dot{A}$. Evaluate the above for $t=0$ to get ${\rm d}(\det)_A(\dot{A}) = {\rm tr}(A^{-1}\dot{A}),$for all $\dot{A}$. Put $\dot{A} = A$ to get ${\rm d}(\det)_A(A) = n \neq 0$. This concludes the proof.

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Proof of formula: in general, if $V$ is a vector space and $F \in {\rm End}(V)$, then we have that $$\begin{align}(\det F)\,\omega(e_1,\ldots,e_n) &= \omega(Fe_1,\ldots,Fe_n), \\ ({\rm tr} \,F)\,\omega(e_1,\ldots, e_n) &= \sum_i \omega(e_1,\ldots, e_{i-1},Fe_i,e_{i+1},\ldots, e_n).\end{align}$$for all determinant forms $\omega$ on $V$. Consider a curve $F(t)$ in ${\rm End}(V)$ and take the derivative of the first expression to get $$\frac{{\rm d}}{{\rm d}t} \det F(t)\,\omega(e_1,\ldots,e_n) = \sum_i \omega(F(t)e_1,\ldots, F(t)e_{i-1}, \dot{F}(t)e_i,F(t)e_{i+1},\ldots, F(t)e_n).$$Write $\dot{F}(t)e_i = F(t)F(t)^{-1}\dot{F}(t)e_i$ and use both identities together to get $$\frac{{\rm d}}{{\rm d}t} \det F(t) \,\omega(e_1,\ldots,e_n) = \det F(t) \,{\rm tr}(F(t)^{-1}\dot{F}(t)) \,\omega(e_1,\ldots,e_n)$$as wanted.