group operations are smooth in $\text{SL}(n, \mathbb{R})$

Question

I am told the following reason as to why group operations of multiplication and inversion are smooth on $\text{SL}(n, \mathbb{R})$. Multiplication is smooth because the matrix entries of a product matrix $AB$ are polynomials in the entries of $A$ and $B$, and inversion is smooth by Cramer's rule. But this sort of feels like a cop out... is there anything I'm missing?

Answer 1

Why do you feel this is a cop-out? It's a very elegant argument, for matrix inversion especially (it's rather obvious that matrix multiplication was going to be smooth). When you invert a matrix using LU or Gaussian elimination, for instance, you do all of these pivoting operations that sure don't look like they're going to be smooth as some choice coefficients pass through zero. Cramer's rule shows that the coefficients of the matrix inverse are just rational polynomials of the matrix entries, with the denominator non-zero as long as the determinant doesn't vanish.

Answer 2

Well, depends on what your definition of smooth maps between manifolds embedded in $\mathbb{R}^k$ is.

If one is using the definition involving pre and post composing charts, then a priori, the multiplication, which is smooth as a map from $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$, need not restrict to a smooth map $\operatorname{SL}(n,\mathbb{R}) \times \operatorname{SL}(n,\mathbb{R}) \to \operatorname{SL}(n,\mathbb{R})$, especially since $\operatorname{SL}(n,\mathbb{R})$ is not an open subset of $\mathbb{R}^{n^2}$. Similarly, the inverse map $\operatorname{GL}(n,\mathbb{R}) \to \operatorname{GL}(n,\mathbb{R})$ is smooth by Cramer's rule, but some argument needs to be made for the restriction.

On the other hand, if one is only dealing with embedded submanifolds of euclidean space, then for simplicity one often defines smooth maps to be those which are locally restrictions of smooth maps between $\mathbb{R}^k$ of various dimensions. In this case, there is nothing more left to show.

The required arguments could of course also be made in general, showing that these two definitions are equivalent. This would likely involve some form of implicit/inverse function theorem.

Answer 3

Here is an argument why multiplication is smooth:

If you apply the definition of the derivative to a linear map $f: \mathbb R^n \to \mathbb R^n$ you see that $f' = f$. This is because the (Fréchet) derivative at $x$ is defined to be the continuous linear map $A$ such that $\lim_{h \to 0}{\|f(x+h) - f(x) - Ah\|\over \|h\|} = 0$.

It is clear that the map $f_A(B)= AB$ is linear hence $(f_A)' = f_A$ and therefore $f_A$ is smooth.

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For smoothness of taking inverses:

The Fréchet derivative of the map $\varphi: A \mapsto A^{-1}$ is the map $B \mapsto -A^{-1}BA^{-1}$. Since we have smoothness of matrix multiplication the derivative of $\varphi$ is smooth hence $\varphi$ itself is smooth.

Answer 4

The key step in what I perceive to showing this is that smooth maps into a manifold $M$ that land in a smooth submanifold $N$ are also smooth maps into the submanifold. This can be checked in local coordinates, by the following argument. Given the constraint that a point $p$ lies on a $k$-dimensional submanifold $N$ of $\mathbb{R}^n$, some of its coordinates are a smooth function of the remaining $k$ coordinates, so in a neighborhood $U$ of $p$ there is a smooth map that sends every point to $N$ which is the identity on $N \cap U$, and this map is clearly smooth with respect tot he chart for $N$ given by choosing the $k$ distinguished coordinates. So locally, our original map into $N$ is the composition of a smooth map into $U$ and a smooth into $N \cap U$, and is therefore smooth.

In this case, $SL(n, \mathbb{R})$ is the smooth submanifold of $GL(n, \mathbb{R})$, which is open in $\mathbb{R}^{n^2}$ and therefore has a single coordinate chart, and the above applies. However, one can now do this for any closed subgroup of $GL(n, \mathbb{R})$.

It turns out that a subgroup $G \subset GL(n, \mathbb{R})$ is a smooth manifold if and only if it is closed as a subset of $GL(n, \mathbb{R})$. The proof of this essentially consists of constructing the tangent space to the identity by looking at sequences that converge to the identity that have limiting directions and showing that the set of limiting directions corresponds to a vector space $V$, and then showing that for some neighborhood of $0$ in $V$, $X \to e^X$ is a chart for $G$ – this can be done without having to invoke more abstract/"powerful"/seemingly tautological or circular results $($i.e. this is a Lie group so everything is trivial$)$. Once we has a chart near the identity, we can multiply its elements by any $g \in G$ to get a chart near $g$.