$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$From Pierre Deligne's [Notes on spinors](https://www.maths.ed.ac.uk/~v1ranick/papers/deligne.pdf), 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.