Skip to main content
added 23 characters in body
Source Link
Dylan Thurston
  • 10.1k
  • 1
  • 44
  • 66

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) (= SO(n) \text{ if $n$ is odd})\\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

The paper also says something about the $h_n(\mathbb{C})$ case, but I don't think it computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

The paper also says something about the $h_n(\mathbb{C})$ case, but I don't think it computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) (= SO(n) \text{ if $n$ is odd})\\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

The paper also says something about the $h_n(\mathbb{C})$ case, but I don't think it computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

added 7 characters in body
Source Link
Dylan Thurston
  • 10.1k
  • 1
  • 44
  • 66

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

HeThe paper also says something about the $h_n(\mathbb{C})$ case, but I don't think heit computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

He also says something about the $h_n(\mathbb{C})$ case, but I don't think he computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

The paper also says something about the $h_n(\mathbb{C})$ case, but I don't think it computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

Found some older references
Source Link
Dylan Thurston
  • 10.1k
  • 1
  • 44
  • 66

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n)\\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

He also says something about the $h_n(\mathbb{C})$ case, but I don't think he computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n)\\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.

To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is

$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) = SO(n) \\ \mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\ \mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$

The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).

This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.


Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):

Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.

He also says something about the $h_n(\mathbb{C})$ case, but I don't think he computes the automorphism group. This is also treated by Jacobson:

Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.

Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.

The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).

Source Link
Dylan Thurston
  • 10.1k
  • 1
  • 44
  • 66
Loading
Post Made Community Wiki by Dylan Thurston