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A bit of context. I am reading about differentiable manifolds and the author gives examples, including GL, OL, SO and SL and he then proves SO is compact and connected but doesn't do the same about GL or SL and I thought a bit of it and have the following ideas. Essentialy, I believe GL can be split into positive and negative determinant and each one of these is like the union of homotheties of SL, which makes me think of SL as a "sphere", so I want to show it is connected to prove these components of GL are connected.

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Consider the linear group of real linear functions $\mathbf{GL}(\mathbf{R}^d) = \mathbf{GL}.$ Define $\mathbf{GL}_\pm$ as those linear functions whose determinant is $\pm,$ accordingly. We endow $\mathbf{GL}$ with the structure of Lie group. Define $\mathbf{SL}$ to be the kernel of the determinant, the latter being an (analytic) diffeomorphism makes of $\mathbf{SL}$ a Lie subgroup of $\mathbf{GL}.$ Similarly, $\mathbf{GL}_\pm$ is the inverse image of the open set $\pm (0, \infty)$ by the determinant, making $\mathbf{GL}_\pm$ also a submanifold of $\mathbf{GL}$ (in fact, $\mathbf{GL}_+$ is a Lie subgroup).

I originally thought of this: using (canonical) coordinates, the determinant is a polynomial and the linear functions in $\mathbf{SL}$ satisfy the relation $\det f = 1,$ which shows $\mathbf{SL}$ is a compact (becasue $\det f = 1$ is a closed and bounded subset of $\mathbf{R}^{d^2}$). However, the comment below says $\mathbf{SL}$ is not compact. What is wrong with this argument?

How to show $\mathbf{SL}$ is connected?

Consider the function $m:\mathbf{R}_+^* \times \mathbf{SL} \to \mathbf{GL}_+$ given according to the rule $m(t, f) = tf.$ Clearly, $m$ is an action that is differentiable (I believe it is even analytic, but I don't care). Since $m$ is surjective (for if $f \in \mathbf{GL},$ then $t f$ has determinant $t^d \det f$ and $t$ can be chosen to make $t^d \det f = 1$), it would turn out $\mathbf{GL}_+$ is connected having the connectedness of $\mathbf{SL}.$

William M.
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  • $\mathbf{SL}$ is not compact, and $m$ is ill-defined. – elidiot Apr 29 '19 at 07:53
  • @elidiot Ammended, I wrote the question with the right $m$ and for some reason I changed it the wrong $m.$ My silly mistake. – William M. Apr 29 '19 at 16:30
  • Also, $\det : {\bf GL} \to {\bf R}^*$ is certainly not an analytic diffeomorphism (for $d > 1$), as it has rank $1 < d^2 = \dim {\bf GL}$. It is, however, a constant rank map, which suffices for your purposes. – Travis Willse Apr 29 '19 at 16:42
  • Ok, thank you. In regards to compactness, I still need to think. I now realise that $diag(1/t,t)$ is a linear homeomorphism of $\mathbf{R}^2$ and its norm can be arbitrarily large, hence $SL$ is not compact. But I will try to understand why my argument breaks. Thanks for the help, the other answer suffices for me. (As a final note, I usually do more measure theory but since I know derivatives very well, I decided to understand more than my comfort zone.) – William M. Apr 29 '19 at 16:45
  • You could elaborate your argument showing that $\bf SL$ is compact in a separate question, and ask where the argument fails. If you do so, include a link to this question for context. – Travis Willse Apr 29 '19 at 17:29
  • The continuous image of a compact set is compact, but the preimage of a compact set doesn't have to be compact. – Jason DeVito - on hiatus Apr 29 '19 at 18:24
  • @JasonDeVito I know what's going on. In d=1 the preimage of a point by a polynomial is a finite set but in greater dimensions they are in fact differential subvarieties of the Euclidean space. My thoughts completely wrong. – William M. Apr 29 '19 at 19:47
  • @Travis I don't understand the comment about the determinant. The determinant is a polynomial, hence analytic (analytic signifying power series expansion and not differentiable as many mathematicians wrongly tend to define). – William M. Apr 29 '19 at 19:48
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    I think Travis was objecting to "diffeomorphism", which typically means a smooth bijection (perhaps, onto its image) of constant rank. – Jason DeVito - on hiatus Apr 29 '19 at 20:04
  • Absolutely, determinant is not a diffeomorphism, it's a morphism of varieties. – William M. Apr 29 '19 at 20:07
  • @WillM. Yes, my objection is just what Jason said, re "...the kernel of the determinant, the latter being an (analytic) diffeomorphism...". – Travis Willse Apr 29 '19 at 20:23

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