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I have that $SL(n,\mathbb{R})$ is an embedded submanifold of dimension $n^2-1$ in $GL(n,\mathbb{R})$, and I know that $T_XGL(n,\mathbb{R})$ is isomorphic to $M(n,\mathbb{R})$ for all $X \in GL(n , \mathbb R)$.

Is there a way I can use this to show that $SL(n,\mathbb{R})$ is a smooth submanifold of $M(n,\mathbb{R})$ and get is dimension? Otherwise, how could I go about it?

Thanks!

mj_indefinite
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  • If $A$ is an embedded submanifold of $B$ and $B$ is an embedded submanifold of $C$, then $A$ is an embedded submanifold of $C$; this is easy to see using the slice chart characterization of such manifolds. The dimension of a manifold is an intrinsic property, i.e., it does not depend on embedding. – Travis Willse May 23 '15 at 09:05

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There is a general theorem of Cartan that a closed subset of a Lie group that is moreover a subgroup is in fact a closed submanifold, so a Lie subgroup.However, this case can be done with the theorem of implicit functions. $SL(n,\mathbb{R})$ is the set of all matrices $(x_{ij})$ with $\det(x_{ij})=1$. Every point of the hypersurface $\det(x_{ij})=1$ is nonsingular. This is a gemeral fact for level sets $P^{-1}(a)$ of homogenous polynomials $P$ for value $a \ne 0$. If at some point $x$ all the partial derivatives $\frac{\partial P}{\partial x_i}(x)=0$ then, using Euler's identity we get $d \cdot P(x) = \sum x_i \frac{\partial P}{\partial x_i}(x) =0$ so $P(x) \ne a$.

Hence, $SL(n,\mathbb{R})$ is a closed submanifold of $M(n,\mathbb{R})$ of dimension $n^2 -1$.

orangeskid
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  • It seems that the OP is asking why $SL(n , \mathbb R)$ is a submanifold of $M(n, \mathbb R)$, not $GL(n, \mathbb R)$ (The OP is aware of the result you mentioned (seem so)) –  May 23 '15 at 09:46
  • @John: Changed the answer a bit. Thanks. – orangeskid May 23 '15 at 09:52