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I was reading Hall's book on Lie groups. After defining Connected Lie groups he stated and proved a proposition :

If $G$ is a matrix Lie group then the component of $G$ containing identity is a subgroup of $G$.

For proving the proposition first he proves that if $A$ and $B$ belongs to the component (which contains identity $I$) we can take the products of the continuous paths connecting $I$ to $A$ and $I$ to $B$ in such a way that the new path will be from$I$ to $AB$.Thus there exists a continuous path from $I$ to $AB$.So $AB$ lies in that component.

Next he has to prove that if $A \in $ component then $A^{-1}$ also lies in that component.For that he argues like this : Since $A$ belongs to this component we can find a continuous path from $I$ to $A$.Let it be $A(t)$.Then $A(t)^{-1}$ is a continuous path from $A^{-1}$ to $I$.This argument I could not understand because

according to my knowledge continuous path $A(t)$ means a continuous function $A(t)$ from $[0,1]$ to $G$ (in this case $A(0)=I $ and $A(1)=A$).

Then $A(t)^{-1}$ will be a map from $G$ to $[0,1]$ such that $A^{-1}(I)=0$ and $A^{-1}(A)=1$. It will not connect $I$ and $A^{-1}$.So how to handle this problem ?

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    $\begingroup$ I think he mean another path $:t \mapsto (A(t))^{-1}$. $\endgroup$ – user99914 Apr 7 '15 at 2:35
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    $\begingroup$ Clarifying what John said: The inverse is not in the sense of inverse function that gives the path, it is the composition of the path function from $[0,1]$ to $G$ followed by the inversion function in the group $G$ which is continuous. (Also note that inverse in the sense of function need not exist as a path is not expected to be a 1-1 function). $\endgroup$ – P Vanchinathan Apr 7 '15 at 2:49
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I think you're just confusing the notation of inverse for a matrix versus inverse of a mapping. Recall that in a Lie group $G$ and given $A \in G$ the map $A \to A^{-1}$ is differentiable. To clarify the notation, let $f:[0,1] \to G$ be differentiable such that $f(0) = I$ and $f(1) = A$. Then take the inverse $A^{-1}$ and define $g(t) = A^{-1}f(t)$. This gives a differentiable map starting at $A^{-1}$ and ending at $I$.

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  • $\begingroup$ Thanks. I was thinking of a path which is inverse function of the previous path. $\endgroup$ – Madhu Apr 7 '15 at 13:16

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