Skip to main content
MathOverflow

Return to Question

added 8 characters in body
Source Link
Марина Marina S
Марина Marina S
  • 317
  • 1
  • 5

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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 $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$$$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. \tag{4} $$

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.

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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 $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$

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.

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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 $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. \tag{4} $$

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.

added 4 characters in body
Source Link
Марина Marina S
Марина Marina S
  • 317
  • 1
  • 5

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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 $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$

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.

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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}. \eqref{2} and \eqref{3} are related by $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$

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.

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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 $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$

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.

`\eqref` and `\DeclareMathOperator`
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

From$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$From Pierre DeligneDeligne's Notes on spinors, we can see that there is an injective group homomorphism (embedding): $$ Spin(2n) \to SU(2^{n-1}) \tag{1} $$\begin{equation} \Spin(2n) \to \SU(2^{n-1}) \label1\tag1 \end{equation} thus the embedding $Spin(2n) \subset SU(2^{n-1})$$\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$$Z(\Spin(4k))=\mathbf{Z}/2 \oplus \mathbf{Z}/2$. When $n=2k+1$, the center $Z(Spin(4k+2))=\mathbf{Z}/4$$Z(\Spin(4k+2))=\mathbf{Z}/4$.

So it is natural to generalize eq (\eqref{1)} to have an injective group homomorphism (embedding), $$ \frac{Spin(4k+2)\times U(1)}{\mathbf{Z}/4} \to \frac{SU(2^{2k})\times U(1)}{\mathbf{Z}/4}. \tag{1'} $$$$ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} \to \frac{\SU(2^{2k})\times \U(1)}{\mathbf{Z}/4}. \label{1'}\tag{1'} $$ The right hand side is true as long as ${\mathbf{Z}/4}\subset Z(SU(2^{2k}))={\mathbf{Z}/2^{2k}}$${\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. (\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), $$ \frac{Spin(4k+2)\times U(1)}{\mathbf{Z}/4} \to \frac{SU(2^{2k})\times U(1)}{\mathbf{Z}/2^{2k}}=U(2^{2k}), \tag{2} $$$$ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} \to \frac{\SU(2^{2k})\times \U(1)}{\mathbf{Z}/2^{2k}}=\U(2^{2k}), \label2\tag2 $$ because the irreducible spinorial representation of $Spin(4k+2)$$\Spin(4k+2)$ is $2^{2k}$ diemsnionaldimensional, which is the same as the standard fundamental representation of $U(2^{2k})$$\U(2^{2k})$. Do you also agree with me on eq (\eqref{2)}?

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

Note for some background: The above we concern the $Spin(2n)=Spin(2n;\mathbb{R})$$\Spin(2n)=\Spin(2n;\mathbb{R})$ acting on the vector space with the real $\mathbb{R}$ field. It is sufficient to study complex representations $\rho: Spin(2n; {\mathbb C})\to SL(2^{n-1}, {\mathbb C})$$\rho: \Spin(2n; {\mathbb C})\to \SL(2^{n-1}, {\mathbb C})$ or $SL(2^{n}, {\mathbb C})$$\SL(2^{n}, {\mathbb C})$ of complex Spin groups $Spin(2n, {\mathbb C})$$\Spin(2n, {\mathbb C})$: The restriction of such a representation to the compact Spin subgroup $Spin(2n)$$\Spin(2n)$ is automatically unitarizable, i.e. the image is contained in a conjugate of $SU(N)$$\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})$$\SO(2n, {\mathbb C})$ (equivalently, $\rho$ is injective). There are also half-spin or semi-spin representations: They are also spinoral. This is related to the above choice of $SL(2^{n-1}, {\mathbb C})$$\SL(2^{n-1}, {\mathbb C})$ for irreducible semi-spin representation, and the above choice of $SL(2^{n}, {\mathbb C})$$\SL(2^{n}, {\mathbb C})$ for reducible spin representation.

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.

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$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}. \eqref{2} and \eqref{3} are related by $$ \begin{array}{ccc} \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/2} & & \\ \downarrow &\searrow & \\ \frac{\Spin(4k+2)\times \U(1)}{\mathbf{Z}/4} & \longrightarrow & \U(2^{2k}) \end{array}. $$

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.

edited title
Link
Марина Marina S
Марина Marina S
  • 317
  • 1
  • 5
Source Link
Марина Marina S
Марина Marina S
  • 317
  • 1
  • 5