Double cover of $GL(n,\mathbb{C})$.

Question

Is there any reference or simple proof to show that the group $\frac{\mathbb{C}\times SL(n,\mathbb{C})}{2\mathbb{Z}}$ is double cover of $GL(n,\mathbb{C})$.

Because $\frac{\mathbb{C}\times SL(n,\mathbb{C})}{2\mathbb{Z}}$ is metalinear group — , Jan 03 '14 at 17:22
I have read somewhere the metalinear group is double cover of general linear group — , Jan 03 '14 at 17:24

score 2 · Accepted Answer · edited Mar 17 '21 at 10:26

I'm likely duplicating the proof in Igor's reference. The idea is that there is a homomorphism $GL_n(\mathbb C) \to \mathbb{C}^* \equiv \mathbb{C} \setminus \{0\}$ by taking the determinant. Since $\mathbb{C}^*$ is the punctured plane, it has a canonical double cover and there is an induced canonical double cover of $GL_n(\mathbb C)$.

Moreover, since the determinant's kernel is $SL_n(\mathbb C)$ there is a nice way of expressing this covering space as $SL_n(\mathbb C) \times_{\mathbb Z_2} \mathbb{C}^*$. The map from $SL_n(\mathbb C)$ to the cover is given by inclusion (lifting property). One can recognise $\mathbb{C}^*$ in $GL_n(\mathbb C)$ as the diagonal matrices. In that way $\mathbb{C}^*$ lifts isomorphically to a subgroup of the double cover. The two groups $SL_n(\mathbb C)$ and $\mathbb{C}^*$ meet in two places in this lift, that's the $\mathbb Z_2$ intersection.

score 0 · Answer 2 · answered Jan 03 '14 at 17:26

A proof is given here (section 9.6): http://books.google.com/books?id=nS_5yegroSoC&pg=PA292&lpg=PA292&dq=metalinear+group&source=bl&ots=vgW39hgnSu&sig=xFMxJWOYB9b3iz_c0rPUaS-JfPk&hl=en&sa=X&ei=PfLGUqa9OJK_sQTx3oCADg&ved=0CE8Q6AEwBQ#v=onepage&q=metalinear%20group&f=false

For posterity, the reference is: A User's Guide to Algebraic Topology By C.T. Dodson, P.E. Parker

Double cover of $GL(n,\mathbb{C})$.

2 Answers2

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