From Pierre Deligne Notes on spinors, we can see that there is an injective group homomorphism (embedding): $$ Spin(2n) \to SU(2^{n-1}) \tag{1} $$ 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 eq (1) to have an injective group homomorphism (embedding), $$ \frac{Spin(4k+2)\times U(1)}{\mathbf{Z}/4} \to \frac{SU(2^{2k})\times U(1)}{\mathbf{Z}/4}. \tag{1'} $$ 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 eq. (1) and (1') are correct for $n \geq 5$? where $4k+2 \geq 10$?

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

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

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 $\rho$ 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.