Injective group homomorphism on $\frac{Spin(4k+2)\times U(1)}{\mathbf{Z}/2}$ or $\frac{Spin(4k+2)\times U(1)}{\mathbf{Z}/4}\to U(2^{2k})$

Question

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$From Pierre Deligne's Notes on spinors, we can see that there is an injective group homomorphism (embedding): \begin{equation} \Spin(2n) \to \SU(2^{n-1}) \label1\tag1 \end{equation} thus the embedding $\Spin(2n) \subset \SU(2^{n-1})$ for some positive integer $n$. I think this seems to be true for $n \geq 5$ but may not be true for $n \leq 4$.

Thus, below let us focus on $n \geq 5$.

We know that when $n=2k$, the center $Z(\Spin(4k))=\mathbf{Z}/2 \oplus \mathbf{Z}/2$. When $n=2k+1$, the center $Z(\Spin(4k+2))=\mathbf{Z}/4$.

So it is natural to generalize \eqref{1} to have an injective group homomorphism (embedding), $$ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} \to \frac{\SU(2^{2k})\times \U(1)}{\mathbf{Z}/4}. \label{1'}\tag{1'} $$ The right hand side is true as long as ${\mathbf{Z}/4}\subset Z(\SU(2^{2k}))={\mathbf{Z}/2^{2k}}$ in the subgroup of center.

My question:

1. Do you agree with me that \eqref{1} and \eqref{1'} are correct for $n \geq 5$? where $4k+2 \geq 10$?

2. I believe that there is an injective group homomorphism (embedding), $$ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} \to \frac{\SU(2^{2k})\times \U(1)}{\mathbf{Z}/2^{2k}}=\U(2^{2k}), \label2\tag2 $$ because the irreducible spinorial representation of $\Spin(4k+2)$ is $2^{2k}$ dimensional, which is the same as the standard fundamental representation of $\U(2^{2k})$. Do you also agree with me on \eqref{2}?

3. Is there an injective group homomorphism (embedding), $$ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} \to \frac{\SU(2^{2k})\times \U(1)}{\mathbf{Z}/2^{2k}}=\U(2^{2k}). \label3\tag3 $$ Here the $\mathbf{Z}/2$ in the denominator is a normal subgroup of $\mathbf{Z}/4$ in \eqref{2}. The \eqref{2} and \eqref{3} are related by $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. \tag{4} $$

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Note for some background: The above we concern the $\Spin(2n)=\Spin(2n;\mathbb{R})$ acting on the vector space with the real $\mathbb{R}$ field. It is sufficient to study complex representations $\rho: \Spin(2n; {\mathbb C})\to \SL(2^{n-1}, {\mathbb C})$ or $\SL(2^{n}, {\mathbb C})$ of complex Spin groups $\Spin(2n, {\mathbb C})$: The restriction of such a representation to the compact Spin subgroup $\Spin(2n)$ is automatically unitarizable, i.e. the image is contained in a conjugate of $\SU(N)$ for appropriate $N$.

A representation $\rho$ of $\Spin(2n, \mathbb C)$ is called spinoral if it does not descend to the orthogonal group $\SO(2n, {\mathbb C})$ (equivalently, $\rho$ is injective). There are also half-spin or semi-spin representations: They are also spinoral. This is related to the above choice of $\SL(2^{n-1}, {\mathbb C})$ for irreducible semi-spin representation, and the above choice of $\SL(2^{n}, {\mathbb C})$ for reducible spin representation.

Answer 1

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 $n=0$.)