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I had read several discussions on these posts on the special unitary $SU(8) \subset SL(8,\mathbb{C})$, also $SO(8) \subset SL(8,\mathbb{R})$, and $Spin(8) \subset Spin(8,\mathbb{C})$.

  1. This post " Embed a Spin group to a special unitary group " by @Moishe Kohan says that the $$SU(n) \supset Spin(N), \tag{1}$$ for the case of the even $N$, there is a half-spin or semi-spin representation, $$N=dim(V(\omega_{\ell}))= 2^{\ell-1}, n=2\ell$$. So according to them, the $n=8$ gives $N=8$. So $SU(8) \supset Spin(8)$.

  2. This post " https://mathoverflow.net/q/295711 " by @David E Speyer seems to conclude that $$SU(n) \supset Spin(n), \quad n \leq 6$$ but $$SU(n) \not \supset Spin(n), \quad n \geq 7.$$ Thus $$SU(8) \not \supset Spin(8). \tag{2}$$

  3. This post " Relation between $SU(8)$, and $Spin(8)$ and $SO(8)/(\mathbf{Z}/2)$ " by @Marc van Leeuwen seems to conclude that $$SU(8) \not \supset Spin(8) \tag{2'} $$ also obviously $$SU(8) \supset SO(8)\simeq Spin(8)/\mathbf{Z}/2 ,$$ but $$SU(8) \not \supset SO(8)/(\mathbf{Z}/2) .$$

Question 1 - Obviously eq (1) contradicts with eq (2) and (2'). I would like to know why there is a contradiction on the result between the 1st post "https://math.stackexchange.com/q/3296240/955245" and other two posts. Where should the modification or correction go?

Question 2 - If we change the special unitary to the unitary group, does that $U(8) \not \supset Spin(8)$ or $U(8) \supset Spin(8)$? Here $U(8) \subset GL(8,\mathbb{C})$.

Please correct me if I said something wrong or imprecise. I am a new member so hopefully I can learn more from your feedback! Thanks!

Marc van Leeuwen
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Марина Marina S
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    Basically the answer is that $Spin(8)$ cannot be embedded as a subgroup of $SU(8)$. Moishe's answers presents a convincing lower bound on N such that $Spin(n)$ embeds into $SU(N)$ but it doesn't prove that it works for this lower bound. In the $n = 8$ (and thus $N=8$) case we can see that it fails (as per David's answer in the second link and Torsten's comment in the first) and the map must have a non-trivial kernel. Indeed it is suggested that this would be a problem whenever n is a multiple of 4. – Callum Aug 04 '21 at 14:14
  • Thanks for the summary, it counts as an answer (almost). except the $U(8)$ part. in Question 2. – Марина Marina S Aug 04 '21 at 16:52
  • At this point my knowledge runs out but I think the answer is no as $Spin(8)$ admits no irreducible faithful representations (according to the discussions in the linked posts while $U(8)$ does. – Callum Aug 05 '21 at 10:36
  • Since it was supposed to summarise my own conclusion (elsewhere), I took the liberty to edit the last two equations so that they no longer contradict each other. – Marc van Leeuwen Aug 05 '21 at 10:46

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