Multiplication map on $SL(n, \mathbb{R})$ is $C^\infty$

Question

In Example 11.16 in Tu's Introduction to Manifolds he shows that the multiplication map of $SL(n, \mathbb{R})$ is smooth. He begins by saying

The multiplication map $$\mu: GL(n, \mathbb{R}) \times GL(n, \mathbb{R}) \rightarrow GL(n, \mathbb{R})\\ (A,B) \mapsto AB,$$ is clearly $C^\infty$ because $$(AB)_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$$ is a polynomial and hence a $C^\infty$ function of the coordinates $a_{ik}$ and $b_{kj}$. However, one cannot conclude in the same way that the multiplication map $$\bar{\mu}: SL(n, \mathbb{R}) \times SL(n, \mathbb{R}) \rightarrow SL(n, \mathbb{R})$$ is $C^\infty$. This is because $\{a_{ij}\}_{1 \leq i, j \leq n}$ is not a coordinate system on $SL(n, \mathbb{R})$; there is one coordinate too many.

And so instead he proves the proposition using another method. However, I am confused on his reasoning as to why we cannot use the polynomial argument to show $SL(n, \mathbb{R})$ is smooth. What does he mean?

Answer 1

Let $M(n,\mathbb R)$ denote the vector space of all real $(n \times n)$-matrices. We can identify $M(n,\mathbb R)$ with $\mathbb R^{n^2}$. The multplication map $\mu$ actually lives on $M(n,\mathbb R)$ and we have $$\mu : M(n,\mathbb R) \times M(n,\mathbb R) \to M(n,\mathbb R), \mu(A,B)_{ij} = (AB)_{ij} = \sum_{k=1}^n a_{ik}b_{kj} .$$ All coordinate functions $\mu(A,B)_{ij}$ are polynomials in the coefficients of $A$ and $B$, hence they are $C^\infty$ and so is $\mu$.

Since $GL(n,\mathbb R)$ is an open subset of $M(n,\mathbb R)$, we see that $$\mu : GL(n,\mathbb R) \times GL(n,\mathbb R) \to GL(n,\mathbb R)$$ is $C^\infty$.

$SL(n,\mathbb R) \subset GL(n,\mathbb R)$ is not open in $M(n,\mathbb R)$, it is a submanifold of $M(n,\mathbb R)$ (and also of $GL(n,\mathbb R)$) having codimension $1$. To verify that $\mu$ is $C^\infty$ on $SL(n,\mathbb R)$ we can

• either work with charts (which are not explictly given here; $SL(n,\mathbb R)$ is known to be submanifold because it is the preimage $\det^{-1}(0)$ of the regular value $0$ of the $C^\infty$-function $\det: M(n,\mathbb R) \to \mathbb R$)
• or use Theorems 11.14 and 11.15 as Tu does.

Working with charts is very complicated, but certainly it does not give a polynomial representation of $\mu$ in terms of local coordinates.