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We know that the spin group embedding $Spin(n) \subset Spin(n+m)$ for any positive integer $n,m$.

Can the universal cover $Spin(n)$ be embedded into some $SO(n+m)$ for appropriate positive integer $n,m$? $$ \begin{array}{ccc} & & Spin(n+m)\\ &\nearrow & \downarrow\\ Spin(n) & \longrightarrow^? & SO(n+m) \end{array} $$

The question is about whether the $?$ map can be an embedding for appropriate positive integer $n,m$? What would be the $n,m$?

p.s. I am interested in $n \geq 3$. Of course $n=2$, we have rather obvious examples...

Марина Marina S
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    Do you want that diagram to commute? Because as long as the image of the diagonal map does not intersect the centre of Spin, you can just take the composition to obtain an embedding, no? – David A. Craven Aug 17 '21 at 18:08
  • I only care the embedding - not sure how that affects the story? please illuminate? – Марина Marina S Aug 17 '21 at 18:09
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    Then just take any map where the centre of Spin(n) is not mapped to the centre of Spin(n+m) and done. – David A. Craven Aug 17 '21 at 18:10
  • sorry but is that an embedding? or just a group homomorphism? I want the embedding... thanks – Марина Marina S Aug 17 '21 at 18:11
  • The composition is injective unless the image of the first map intersects the kernel of the second map. This is just basic function theory.. – David A. Craven Aug 17 '21 at 18:14
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    @DavidA.Craven: Of course, one has to prove that you can find such a map $Spin(n)\rightarrow Spin(n+m)$. As a hint to the OP: if you can find a faithful rep of $Spin(n)$, this gives you an embedding $Spin(n)\rightarrow U(N)$ for some possibly large $N$. Can you then find an imbedding of $U(N)$ into $SO(n+m)$ for $m$ large enough? – Jason DeVito - on hiatus Aug 17 '21 at 18:25
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    @JasonDeVito The OP already thinks they have a map from Spin to a larger Spin. Depending on this map, they might already have an answer. – David A. Craven Aug 17 '21 at 18:29
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    @DavidA.Craven: Agreed. However, at least when $m=1$, for $n\geq 8$, the only embeddings share centers, so it may require some more thought by the OP. – Jason DeVito - on hiatus Aug 17 '21 at 18:33
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    @JasonDeVito Absolutely. $m$ will have to grow fairly quickly. – David A. Craven Aug 17 '21 at 18:43
  • There is some answer from n embedding ()→() https://math.stackexchange.com/a/3296484/955245. With this inform can some or all of you contribute answers? many thanks - I want to see solutions explicitly. – Марина Marina S Aug 17 '21 at 22:49
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    You included a diagram involving $Spin(n+m)$. That makes it seem like you want the embedding $Spin(n) \to SO(n+m)$ to satisfy some additional property, namely that the diagram commutes. Is that what you're asking? If so, you need to specify the map $Spin(n) \to Spin(n+m)$. If it is the standard one, then there is no embedding $Spin(n) \to SO(n+m)$ for which the diagram commutes. If you're free to choose the map $Spin(n)\to Spin(n+m)$, then there may be embeddings $Spin(n) \to SO(n+m)$ for which the diagram commutes. If you don't care about the diagram commuting, then Jason's comment suffices. – Michael Albanese Aug 18 '21 at 20:36

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