Given $G=\left\{A\in M_2(\mathbb{R})\mid A^\top XA = X\right\}$ where $X = \pmatrix{3&1\\1&1}$ and a Lie algebra $\mathfrak g=\left\{Y\in M_2(\mathbb{R})\mid Y^\top X+XY = 0\right\}$, how would I show that every element in $G$ does not have $\det=-1$?
Finding determinant of matrix lie group
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lie-algebras
lie-groups
1 Answers
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The matrix $ A = \pmatrix{0&\sqrt{3}\\\frac{1}{\sqrt3}&0} $ has $\det{A} = - 1$ and satisfies $A^T X A = X$ unless I miscalculated....
(The Group in question can be thought of as the orthogonal group relative to the scalar product induced by $X$, which has two components (in the sense of topology), like the usual $O(2)$. These are precisely the counterimages of $1, -1$ respectively, under $\det$)
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0$G$, if defined as in your post, _does_ have two components, they won't go away. If you are asked to show that only one component exists an additional condition has to be imposed. That said, your original question does not involve the Lie Algebra in any way. What is true (as a possible additional condition) is that the image of this Lie algebra under the exponential map at $E$ is a connected Lie sub group of $G$ (namely the one for which $\det = 1$). – 2011-12-31