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Inspired by a previous post, Embed a Spin group to a special unitary group

I am wondering what are the magic properties of the quotient space from $$U(2^{l-1})/{\rm Spin}(2 l)$$ that makes such an embedding possible: $${\rm Spin}(2 l) \subset U(2^{l-1}).$$

For example, I know that ${\rm Spin}(k+1)/{\rm Spin}(k)=S^k$ is a $k$-sphere.

  • Do we have similar properties to describe the quotient space from $U(2^{l-1})/{\rm Spin}(2 l)$?

  • What are the homotopy groups of this space? $\pi_d(U(2^{l-1})/ {\rm Spin}(2 l))=?$

Any references are welcome.

annie marie cœur
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  • What exactly do you mean by $U(n, {\mathbb C})$? The standard meaning would be the complexification of $U(n)$, which is usually denoted $GL(n, {\mathbb C})$. – Moishe Kohan Apr 05 '20 at 01:47
  • I meant $U(2^{l-1}, {\mathbb C})=U(2^{l-1})$. thanks! – annie marie cœur Apr 05 '20 at 02:15
  • OK, then what do you mean by $Spin(n, {\mathbb C})$? Usually this notation means a complex Lie group. It cannot possibly embed in a compact Lie group such as a unitary group. I suspect you really did not mean to write ${\mathbb C}$ anywhere in your question and $Spin$ stands for a compact group. – Moishe Kohan Apr 05 '20 at 02:30
  • thanks! but I follow the notation from the expert math.stackexchange.com/questions/3296240/ – annie marie cœur Apr 05 '20 at 03:19
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    You should not follow the notation blindly. There are two different groups, namely the real Spin group which is compact and the complex Spin group which is noncompact. In my answer I was working with complex Spin groups which makes sense in the context of my answer. In your case, you cannot work with the complex Spin groups. – Moishe Kohan Apr 05 '20 at 03:22
  • @Moishe Kohan changed accordingly, thanks – annie marie cœur Apr 05 '20 at 03:25

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