4

We knew that

also

I would like to generalize the result.

Question: I hope to prove or disprove that

$$ SU(n)\times Spin(k) \subset \frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}? $$

Does this hold for general $n$ or $k$? or some range of $n\geq 2,k\geq 3$?

here $\mathbb{Z}/2$ is the order 2 normal subgroup of ${Spin}(2n)$ also the order 2 normal subgroup of $ Spin(k)$. So we mod out the common $\mathbb{Z}/2$. Namely, we consider a more exotic $\frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}\neq {SO}(2n)\times Spin(k)$ and $\frac{{Spin}(2n)\times Spin(k)}{\mathbb{Z}/2}\neq {Spin}(2n)\times SO(k)$, we do not consider the product of two gruops.

I check that for lower dimensions $n\le 6$ there are accidental isomorphisms of spin groups with other familiar Lie groups: $\DeclareMathOperator{Spin}{\mathrm{Spin}}$

$$\begin{array}{|l|l|} \hline Spin(1) & {O}(1)=\mathbb{Z}/2 \\ \hline Spin(2) & {SO}(2)=U(1) \\ \hline Spin(3) & {Sp}(1)=SU(2) \\ \hline Spin(4) & {Sp}(1)\times{Sp}(1)=SU(2) \times SU(2) \\ \hline Spin(5) & {Sp}(2) \\ \hline Spin(6) & {SU}(4) \\ \hline \end{array} $$ Here I list down the Lie group isomorphisms, not the Lie algebra.

So we can check: $n=2,k=3$, $$ SU(2)\times Spin(3) \subset \frac{{Spin}(4)\times Spin(3)}{\mathbb{Z}/2}= \frac{SU(2)\times SU(2)\times Spin(3)}{\mathbb{Z}/2} \text{ is true}? $$ How about general $n\geq 2,k\geq 3$?

annie marie cœur
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1 Answers1

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First, the center of $Spin(4k+2)$ is a $\mathbb{Z}_4$, and the center of $Spin(odd)$ is a $\mathbb{Z}_2$, so there is a unique $\mathbb{Z}_2$ you can quotient by in each of these cases. But for $Spin(4k)$, the center is $\mathbb{Z}_2\oplus \mathbb{Z}_2$, so there is ambiguity when you say "the order two normal subgroup of $Spin(4k)$". I'll assume you're referring to the $\mathbb{Z}_2$ for which $Spin(4n+2)/\mathbb{Z}_2 \cong SO(4n+2)$.

With that out of the way, we claim that:

For any $n\geq 2$ and $k\geq 3$, there is a Lie subgroup of $G:= (Spin(2n)\times Spin(k))/\mathbb{Z}_2$ which is isomorphic to $H:=SU(n)\times Spin(k)$.

Before proving this, recall that in your previous question, I showed that there is an $SU(n)$ in $Spin(2n)$ obtained by lifting the $SU(n)$ in $SO(2n)$. It follows that if $p:Spin(2n)\rightarrow SO(2n)$ is the double cover map, then $p|_{SU(n)}$ is injective.

Now, let's prove the theorem.

Proof. Let $\overline{G} = Spin(2n)\times Spin(k)$ be the universal cover of $G$, and suppose $\pi:\overline{G}\rightarrow G$ is the double covering map with kernel given by $\mathbb{Z}_2 = \langle z\rangle$.

In your previous question, we already established that there is a subgroup of $Spin(2n)$ isomorphic to $SU(n)$. Now, let $\overline{H} = SU(n)\times Spin(k)\subseteq Spin(2n)\times Spin(k)$. Obviously, $\overline{H}$ is isomorphic to $H$.

Consider $\pi(\overline{H})\subseteq G$. We claim that $\pi(\overline{H})$ is isomorphic to $H$. To that end, note that $\pi|_{\overline{H}}:\overline{H}\rightarrow \pi(\overline{H})$ is a surjective Lie homomorphism by definition, so we just need to check that it is injective, and then we'll have the isomorphisms $H\cong \overline{H}\cong \pi(\overline{H})$.

Since $\ker \pi_{\overline{H}}\subseteq \ker \pi$, we need only show $z\notin \ker \pi|_{\overline{H}}$. So, assume for a contradiction that $z\in \ker \pi|_{\overline{H}}\subseteq \overline{H}$. Letting $\rho:\overline{G}\rightarrow Spin(2n)$ be the projection map, we see that $\rho(z)\in \rho(\overline{H}) = SU(n)\subseteq Spin(2n)$.

But the projection map $p:Spin(2n)\rightarrow SO(2n)$ is simply $Spin(2n)\rightarrow Spin(2n)/\langle\rho(z)\rangle \cong SO(2n)$. If $\rho(z)\in SU(n)\subseteq Spin(2n)$, then the projection map $p$ restricted to $SU(n)$ is not injective, contradicting the fact mentioned before the proof. This establishes that $z\notin \ker\pi_{\overline{H}}$, so $\pi_{\overline{H}}$ is a Lie isomorphism from $\overline{H}$ to $\pi(\overline{H})$. $\square$.

Jason DeVito - on hiatus
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  • Thanks very much, voted up, I am particularly curious about the case for $Spin(4k+2)$. As you clarify nicely, the center of $Spin(4k+2)$ is a $Z(Spin(4k+2))=\mathbb{Z}_4$. The $\mathbb{Z}_2$ to mod out is the normal subgroup $\mathbb{Z}_2 \subset \mathbb{Z}_4 =Z(Spin(4k+2))$ – annie marie cœur Apr 03 '20 at 22:38
  • Excuse me, Is there a typo/mistake here "Obviously, $\bar H$ is isomorphic to ." Do you mean "Obviously, a subgroup of $\bar H$ is isomorphic to ." instead? – annie marie cœur Apr 03 '20 at 22:59
  • $H$ is a group abstractly isomorphic to $SU(n)\times Spin(k)$. The group $\overline{H}$ is a specific subgroup of $\overline{G}$ of the form $SU(n)\times Spin(k)$. I think $H$ and $\overline{H}$ are isomorphic.. – Jason DeVito - on hiatus Apr 03 '20 at 23:04
  • Maybe I was confused by your notation: $\overline{H} \cong H$ by definition. But $\overline{G} \not \cong G$. Yes? – annie marie cœur Apr 03 '20 at 23:29
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    Correct. $ \text{ }$ – Jason DeVito - on hiatus Apr 03 '20 at 23:29
  • So $\pi(\overline{H})=\pi({H})$. Do we need to clarify the meaning of $\pi(\overline{H})=\pi({H})$? Since we are taking the mod $\mathbb{Z}_2$ out of their common subgroup in $\overline{H}=H=SU(n)\times Spin(k)$? The $SU(n) \supset \mathbb{Z}_n$ as the center, only only when $n$ is even we have $SU(n) \supset \mathbb{Z}_n \supseteq \mathbb{Z}_2$ ? Thank you! – annie marie cœur Apr 03 '20 at 23:44
  • I am not sure how we define $\pi(\overline{H})=\pi({H})$ when $SU(n)$ with $n$ odd? – annie marie cœur Apr 03 '20 at 23:47
  • @Jason DeVito: By "Since $\ker \pi_{\overline{H}}\subseteq \ker \pi$," do you mean $\ker \pi |_{\overline{H}}\subseteq \ker \pi$? (just make sure the notation.) – wonderich Apr 04 '20 at 01:01
  • What is the difference between $\ker \pi_{\overline{H}}$ and $\ker \pi|_{\overline{H}}$??? – annie marie cœur Apr 04 '20 at 02:21
  • My rational says that it means "Since $\ker \pi|{\overline{H}}\subseteq \ker \pi:= \ker \pi|{\overline{G}}$. – wonderich Apr 04 '20 at 03:02
  • @Jason DeVito, by this argument, is this true that all positive integer $≥1,≥1$ also hold for the proven relation? – annie marie cœur Apr 04 '20 at 03:42
  • @Annie: $H$ is an abstract Lie group, not presented as a subgroup of anything. In particular, $H\not\subseteq \overline{G}$, so writing $\pi(H)$ doesn't make sense. On the other hand, after we identity a subgroup of $Spin(2n)$ as being $SU(n)$, it makes sense to write $\overline{H} = SU(n)\times Spin(k)\subseteq \overline{G}$. The $\mathbb{Z}_2\in Spin(2n)$ you quotient by to get $SO(2n)$ is not an element of $SU(n)$ for the subgroup we're using. And $\pi$ is defined from $\overline{G}$ to $G$, so whether or not $n$ is even or odd, $\pi(\overline{H})$ makes sense. – Jason DeVito - on hiatus Apr 04 '20 at 21:36
  • @annieheart: For $k\leq 2$, what do you mean by $Spin(2)$? For the other $k$, the map $Spin(k)\rightarrow SO(k)$ is both a universal and a double cover. Is $Spin(2)\rightarrow SO(2)$ a universal or a double cover? Same with $Spin(1)\rightarrow SO(1)$? – Jason DeVito - on hiatus Apr 04 '20 at 21:39
  • $Spin(1)=O(1)=Z_2$ https://en.wikipedia.org/wiki/Spin_group#Accidental_isomorphisms – annie marie cœur Apr 04 '20 at 23:30
  • @annie: I assume you meant $O(1)$, not $SO(1)$? In any event, it seems as though my proof works regardless of what $Spin(k)$ means when $k\leq 2$. – Jason DeVito - on hiatus Apr 04 '20 at 23:35
  • @wonderich: Sorry, $\pi_{\overline{H}}$ is a typo - it should be $\pi|_{\overline{H}}$. – Jason DeVito - on hiatus Apr 04 '20 at 23:35
  • accepted, thanks guys. maybe some of you know this math.stackexchange.com/q/3610249/141334 ? comments welcome! – annie marie cœur Apr 05 '20 at 01:14