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I am told that I need to find a path $c(t)$ such that $c(t)=x(t), X(0)=x \forall X s.t. det X=1$. So I can show that $d/dt(f(c(t))$ at $t=0=[d_{f(c(t))}f](c'(t))]\ne 0$

My problem is how to explicitly construct such a path $c(t)$?

grayQuant
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    It's a single polynomial constraint on the entries, hence... – Adam Hughes Apr 28 '16 at 20:43
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    Just adding to Adam Hughes' comment: If $f:\mathbb{R}^N\to\mathbb{R}^d$ is differentiable and $c$ is a regular value of $f$, then $f^{-1}(c)$ is a submanifold of dimension $N-d$: https://en.wikipedia.org/wiki/Preimage_theorem. Use this with $f$ the determinant (which is polynomial on the entries of matrices) and $c=1$. – Luiz Cordeiro Apr 28 '16 at 20:56
  • @LuizCordeiro still not sure what the path in matrix representation should be – grayQuant Apr 28 '16 at 21:33
  • @grayQuant So the question you are posing is if $SL(n)$ (matrices of determinant $1$) is path connected. This is harder, see http://math.stackexchange.com/questions/315356/prove-that-the-set-of-n-by-n-real-matrices-with-positive-determinant-is-conn. In the answer it is shown that the space of matrices of positive determinant is path connected, but we don't get a precise formula for the paths. You multiply these paths by $\det^{1/n}$ to ensure you get determinant 1. – Luiz Cordeiro Apr 28 '16 at 21:43

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If I understand your question well, what you are looking for is the following: Given a matrix $X$ with $\det(X)=1$ find an explicit curve $c$ with $c(0)=X$ such that $t\mapsto \det(c(t))$ has non-vanishing derivative for $t=0$ (so this shows that $D\det(X)\neq 0$. If this is indeed the question, then you can simply take $c(t)=(1+t)X$, since this has determinant $(1+t)^n$, so its derivative at $t=0$ is $n\neq 0$.

Andreas Cap
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