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Fixed a few typos and infelicitous phrasings.
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Robert Bryant
Robert Bryant
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You really should have a look at F. Reese Harvey's book Spinors and Calibrations, where all of your questions are answered.

For example, your 'inclusion' (1) is not correct for sufficiently large $n$. In fact, here are the actual embeddings:

There is an embedding $\mathrm{Spin}(8n)\hookrightarrow\mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$, but this latter product does not embed into $\mathrm{SU}(2^{4n-1})$, nor does $\mathrm{Spin}(8n)$, each. Each of the two projections into the $\mathrm{SO}(2^{4n-1})$ factors is a double cover of its image, which is a maximal proper subgroup $\mathrm{SO}'(8n)\subset\mathrm{SO}(2^{4n-1})$ that, for $n>1$, is not isomorphic to $\mathrm{SO}(8n)$. Under the above embedding, the center of $\mathrm{Spin}(4n)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$ where $\epsilon_i^2 = 1$. (The case $n=1$ is, of course, the famous triality isomorphism, $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$.)

There is an embedding $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})\subset \mathrm{SU}(2^{4n+1})\times\mathrm{SU}(2^{4n+1})$, but, for $n>0$, the projections of $\mathrm{Spin}(8n{+}4)$ into either $\mathrm{Sp}(2^{4n})$ factor is a double cover onto a maximal proper subgroup $\mathrm{SO}'(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ that is not isomorphic to $\mathrm{SO}(8n{+}4)$. In particular, $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{SU}(2^{4n+1})$ is never an embedding. Under the above embedding, the center of $\mathrm{Spin}(8n{+}4)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})$ where $\epsilon_i^2 = 1$. (The case $n=0$ is, of course, different, because, in this case, the embedding $\mathrm{Spin}(4)\hookrightarrow\mathrm{Sp}(1)\times \mathrm{Sp}(1)=\mathrm{SU}(2)\times\mathrm{SU}(2)$ is an isomorphism.)

Finally, for $n\ge 1$ there is an embedding $\mathrm{Spin}(4n{+}2)\hookrightarrow\mathrm{SU}(2^{2n})$. (Of course, this fails for $n=0$. The image of this embedding is a maximal proper subgroup of $\mathrm{SU}(2^{2n})$, and, under this embedding, the center of $\mathrm{Spin}(4n{+}2)$ goes to the multiples of the identity by a power of $i\in\mathbb{C}$. (Of course, this fails for $n=0$.)

You really should have a look at F. Reese Harvey's book Spinors and Calibrations, where all of your questions are answered.

For example, your 'inclusion' (1) is not correct for sufficiently large $n$. In fact, here are the actual embeddings:

There is an embedding $\mathrm{Spin}(8n)\hookrightarrow\mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$, but this latter product does not embed into $\mathrm{SU}(2^{4n-1})$, nor does $\mathrm{Spin}(8n)$, each of the two projections into the $\mathrm{SO}(2^{4n-1})$ factors is a double cover of its image, which is a maximal proper subgroup $\mathrm{SO}'(8n)\subset\mathrm{SO}(2^{4n-1})$ that, for $n>1$, is not isomorphic to $\mathrm{SO}(8n)$. Under the above embedding, the center of $\mathrm{Spin}(4n)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$ where $\epsilon_i^2 = 1$. (The case $n=1$ is, of course, the famous triality isomorphism, $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$.)

There is an embedding $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})\subset \mathrm{SU}(2^{4n+1})\times\mathrm{SU}(2^{4n+1})$, but, for $n>0$, the projections of $\mathrm{Spin}(8n{+}4)$ into either $\mathrm{Sp}(2^{4n})$ factor is a double cover onto a maximal proper subgroup $\mathrm{SO}'(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ that is not isomorphic to $\mathrm{SO}(8n{+}4)$. In particular, $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{SU}(2^{4n+1})$ is never an embedding. Under the above embedding, the center of $\mathrm{Spin}(8n{+}4)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})$ where $\epsilon_i^2 = 1$. (The case $n=0$ is, of course, different, because, in this case, the embedding $\mathrm{Spin}(4)\hookrightarrow\mathrm{Sp}(1)\times \mathrm{Sp}(1)=\mathrm{SU}(2)\times\mathrm{SU}(2)$ is an isomorphism.)

Finally, for $n\ge 1$ there is an embedding $\mathrm{Spin}(4n{+}2)\hookrightarrow\mathrm{SU}(2^{2n})$. (Of course, this fails for $n=0$. The image of this embedding is a maximal proper subgroup of $\mathrm{SU}(2^{2n})$, and, under this embedding, the center of $\mathrm{Spin}(4n{+}2)$ goes to the multiples of the identity by a power of $i\in\mathbb{C}$.

You really should have a look at F. Reese Harvey's book Spinors and Calibrations, where all of your questions are answered.

For example, your 'inclusion' (1) is not correct for sufficiently large $n$. In fact, here are the actual embeddings:

There is an embedding $\mathrm{Spin}(8n)\hookrightarrow\mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$, but this latter product does not embed into $\mathrm{SU}(2^{4n-1})$, nor does $\mathrm{Spin}(8n)$. Each of the two projections into the $\mathrm{SO}(2^{4n-1})$ factors is a double cover of its image, which is a maximal proper subgroup $\mathrm{SO}'(8n)\subset\mathrm{SO}(2^{4n-1})$ that, for $n>1$, is not isomorphic to $\mathrm{SO}(8n)$. Under the above embedding, the center of $\mathrm{Spin}(4n)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$ where $\epsilon_i^2 = 1$. (The case $n=1$ is, of course, the famous triality isomorphism, $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$.)

There is an embedding $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})\subset \mathrm{SU}(2^{4n+1})\times\mathrm{SU}(2^{4n+1})$, but, for $n>0$, the projections of $\mathrm{Spin}(8n{+}4)$ into either $\mathrm{Sp}(2^{4n})$ factor is a double cover onto a maximal proper subgroup $\mathrm{SO}'(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ that is not isomorphic to $\mathrm{SO}(8n{+}4)$. In particular, $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{SU}(2^{4n+1})$ is never an embedding. Under the above embedding, the center of $\mathrm{Spin}(8n{+}4)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})$ where $\epsilon_i^2 = 1$. (The case $n=0$ is, of course, different, because, in this case, the embedding $\mathrm{Spin}(4)\hookrightarrow\mathrm{Sp}(1)\times \mathrm{Sp}(1)=\mathrm{SU}(2)\times\mathrm{SU}(2)$ is an isomorphism.)

Finally, for $n\ge 1$ there is an embedding $\mathrm{Spin}(4n{+}2)\hookrightarrow\mathrm{SU}(2^{2n})$. The image of this embedding is a maximal proper subgroup of $\mathrm{SU}(2^{2n})$, and, under this embedding, the center of $\mathrm{Spin}(4n{+}2)$ goes to the multiples of the identity by a power of $i\in\mathbb{C}$. (Of course, this fails for $n=0$.)

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Robert Bryant
Robert Bryant
  • 108.4k
  • 8
  • 342
  • 453

You really should have a look at F. Reese Harvey's book Spinors and Calibrations, where all of your questions are answered.

For example, your 'inclusion' (1) is not correct for sufficiently large $n$. In fact, here are the actual embeddings:

There is an embedding $\mathrm{Spin}(8n)\hookrightarrow\mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$, but this latter product does not embed into $\mathrm{SU}(2^{4n-1})$, nor does $\mathrm{Spin}(8n)$, each of the two projections into the $\mathrm{SO}(2^{4n-1})$ factors is a double cover of its image, which is a maximal proper subgroup $\mathrm{SO}'(8n)\subset\mathrm{SO}(2^{4n-1})$ that, for $n>1$, is not isomorphic to $\mathrm{SO}(8n)$. Under the above embedding, the center of $\mathrm{Spin}(4n)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$ where $\epsilon_i^2 = 1$. (The case $n=1$ is, of course, the famous triality isomorphism, $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$.)

There is an embedding $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})\subset \mathrm{SU}(2^{4n+1})\times\mathrm{SU}(2^{4n+1})$, but, for $n>0$, the projections of $\mathrm{Spin}(8n{+}4)$ into either $\mathrm{Sp}(2^{4n})$ factor is a double cover onto a maximal proper subgroup $\mathrm{SO}'(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ that is not isomorphic to $\mathrm{SO}(8n{+}4)$. In particular, $\mathrm{Spin}(8n{+}4)\hookrightarrow\mathrm{SU}(2^{4n+1})$ is never an embedding. Under the above embedding, the center of $\mathrm{Spin}(8n{+}4)$ goes to the subgroup of order $4$ consisting of the elements $(\epsilon_1 I,\epsilon_2 I)\in \mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})$ where $\epsilon_i^2 = 1$. (The case $n=0$ is, of course, different, because, in this case, the embedding $\mathrm{Spin}(4)\hookrightarrow\mathrm{Sp}(1)\times \mathrm{Sp}(1)=\mathrm{SU}(2)\times\mathrm{SU}(2)$ is an isomorphism.)

Finally, for $n\ge 1$ there is an embedding $\mathrm{Spin}(4n{+}2)\hookrightarrow\mathrm{SU}(2^{2n})$. (Of course, this fails for $n=0$. The image of this embedding is a maximal proper subgroup of $\mathrm{SU}(2^{2n})$, and, under this embedding, the center of $\mathrm{Spin}(4n{+}2)$ goes to the multiples of the identity by a power of $i\in\mathbb{C}$.