Showing that the path-connected component of the identity matrix in a subgroup of $GL_n(\Bbb R)$ is a normal subgroup

Question

Let $M(n;\mathbb{R})$ denote the set of all $n \times n$ matrices with real entries (identified with $\mathbb{R}^{n^{2}}$ and endowed with its usual topology) and let $GL(n;\mathbb{R})$ denote the group of invertible matrices. Let $G$ be a subgroup of $GL(n;\mathbb{R})$. Define $$H = \biggl\{ A \in G \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}$$

Then is $H$ normal in $G$?

For proving $H$ normal i should verify two things:

• First $H$ is a subgroup.

• For all $A \in G, B \in H$, we must have $A \cdot B \cdot A^{-1} \in H$.

That's it i am not able to proceed any further.

Answer 1

actually, the answer jug gave in his comments is easier, but here is another proof (though a bit overkill) which shows that this group is actually the matrices with positive determinant.

The function $\det:GL_n(\mathbb{R})\rightarrow \mathbb{R}$ is continuous, so if $\varphi:[0,1]\rightarrow \mathbb{R}$ such that $\varphi(0)=A,\; \varphi(1)=I$ is continuous then $\det \circ \varphi$ is continuous. Since $\det(\varphi(0))=\det(A)$ and $\det\varphi (1)=1$ then we must have $\det(A)>0$ otherwise by the intermediate value theorem there is $t$ such that $\det(\varphi (t))=0$ but then $\varphi(t) \notin GL_n(\mathbb{R})$.

On the other hand, suppose that $\det(A)>0$ then if $E$ is an elementary matrix (corresponding to a row operation) then define $\varphi(t)=((1-t)I+tE)A$ for $0\leq t \leq 1$. It is continuous and its image is in $GL_n(\mathbb{R})$ if $((1-t)I+tE)$ is always invertible. This is for true if E is a multiplication by a positive scalar or if it is an addition of one row to another, but not true for multiplication by a negative scalar, or switching rows.

Show that if $\det(A)>0$ then you can use such elementary matrices to go from $A$ to $I.$

This shows that your $H$ is just all the matrices with positive determinant.

Answer 2

${\rm GL}_n({\bf R})$ has at least two connected components. It is disconnected by the continuous function $f: {\rm GL}_n({\bf R})\rightarrow \{-1, 1\}$ defined by $f(T) = \det(T)/|\det(T)|$. Since ${\rm GL}_n({\bf R})$ is a manifold, it is locally path connected. Therefore, it path components coincide with its connected component.

You can find a detailed proof that the matrices of positive determinant are connected in Lee, J., {\it Introduction to Smooth Manfolds}, on page 236. This requires a little exercise in the analysis of the Graham-Schmidt orthonormalization procedure to produce a path connecting any matrix with positive determinant to the identity.

Switching the first two rows of a matrix defines a homeomorphism of the general linear group onto itself. It maps the matrices of positive determinant homeomorphically onto those of negative determinant. Therefore we have exactly two connected components.

Answer 3

Here's another proof.

Since $Gl = GL_n(\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$, it is naturally a smooth manifold. Hence, the path components are the same as the components of it. Further, since multiplication and inverses can be written as polynomials in the entries of the matrix, these operations are continuous (smooth in fact).

$H$ is clearly the path component of $Gl$, by the above, it's actually a component in the usual sense.

Now, fix $h\in H$ and consider the map $L_h:Gl\rightarrow Gl$ sending $A$ to $hA$. This is continuous by the above, so it sends components to components. Where does it send $H$? Well, since $I\in H$, we see that $L_h I = hI = h$, so $L_h$ sends one point in $H$ to another point in $H$. It follows that $L_h$ sends all of $H$ into itself.

This shows that $H$ is closed under multiplication.

What about inverses?

Well, first note that $L_h$ is a homeomorphism because it has inverse $L_h^{-1} = L_{h^{-1}}$. This shows that the ONLY component it sends onto $H$ is $H$. Let $X$ be the component of $Gl$ containing $h^{-1}$. Then we know $L(X) \subseteq H$ since $h^{-1}\in X$ and $L_h h^{-1} = hh^{-1} = Id\in H$. This implies that $X = H$, i.e., that $h^{-1}\in H$. So, $H$ is closed under inverses.

We have now shown that $H$ is a subgroup - now we need only show it's normal.

So, let $\in Gl$. Consider the map $C_g:Gl\rightarrow Gl$ sending $A$ to $hAh^{-1}$. As above, this map is continuous. So, it sends components to components. Where does it send $H$? Well, $I\in H$ and $C_g(I) = gIg^{-1} = I$, so it sends $H$ to $H$, that is, $C_g(H)\subseteq H$. But $C_g(H) = gHg^{-1}$, so $H$ is normal.

(Incidentally, what I really proved is the following: Let $G$ be any Lie group. Let $G_0$ be the identity path component of $G$. Then $G_0$ is a normal subgroup of $G$. More generally, if $G$ is a topological group such that components and path components agree, this works. I'm not sure whether or not $G$ is a topological group implies the components and path components agree, though).