- #1

- 4

- 0

$$\langle p,q \rangle := G(p,q)+i\Omega^{I}(p,q)+j\Omega^{J}(p,q)+k\Omega^{K}(p,q)$$

really satisfy hyperhermitian condition?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- A
- Thread starter Leditto
- Start date

- #1

- 4

- 0

$$\langle p,q \rangle := G(p,q)+i\Omega^{I}(p,q)+j\Omega^{J}(p,q)+k\Omega^{K}(p,q)$$

really satisfy hyperhermitian condition?

- #2

chiro

Science Advisor

- 4,790

- 132

Could you please (for those of us like myself unfamiliar with the field and terminology) give a description of the condition?

Also - what is a quaternionic structure? I know what a quaternion is - is it just a tensor product of three quaternions?

- #3

Ben Niehoff

Science Advisor

Gold Member

- 1,879

- 162

Could you please (for those of us like myself unfamiliar with the field and terminology) give a description of the condition?

Also - what is a quaternionic structure? I know what a quaternion is - is it just a tensor product of three quaternions?

On a vector space, a quaternionic structure is a set of three linear operators ##I,J,K## such that

$$I^2 = J^2 = K^2 = IJK = -\mathrm{id}, \quad IJ = K, \quad JK = I, \quad KI = J.$$

However, I've not heard the word "hyperhermitian" before.

- #4

jim mcnamara

Mentor

- 4,282

- 2,883

- #5

- 4

- 0

On a vector space, a quaternionic structure is a set of three linear operators ##I,J,K## such that

$$I^2 = J^2 = K^2 = IJK = -\mathrm{id}, \quad IJ = K, \quad JK = I, \quad KI = J.$$

However, I've not heard the word "hyperhermitian" before.

Thanks for your response, Niehoff.

In complex case, Hermitian condition is described by $$\langle I u,I v \rangle=\langle u,v \rangle.$$ Quaternionic analogue of that condition is called hyperhermitian condition and defined by $$\langle I u,I v \rangle=\langle J u,J v \rangle=\langle K u,K v \rangle = \langle u,v \rangle.$$ In addition, there are metric compatibilities condition that make vector space ##V## a hyperkahler manifold, $$G(Iu,v)=\Omega^{I}(u,v),\quad G(Ju,v)=\Omega^{J}(u,v),\quad G(Ku,v)=\Omega^{K}(u,v).$$ I've checked that hyperhermitian condition can't be fulfilled by defining $$\langle u,v \rangle=G(u,v)+i\,\Omega^{I}(u,v)+j\,\Omega^{J}(u,v)+k\,\Omega^{K}(u,v).$$ My calculation:

\begin{eqnarray*}

\langle I u,I v \rangle&=&G(Iu,Iv)+i\,\Omega^{I}(Iu,Iv)+j\,\Omega^{J}(Iu,Iv)+k\,\Omega^{K}(Iu,Iv)\\

&=&\Omega^{I}(u,Iv)+i\,G(I^2u,Iv)+j\,G(JIu,Iv)+k\,G(KIu,Iv)\\

&=&-\Omega^{I}(Iv,u)-i\,G(Iv,u)-j\,G(Ku,Iv)+k\,G(Ju,Iv)\\

&=&-G(I^2v,u)-i\,\Omega^{I}(v,u)-j\,\Omega^{K}(u,Iv)+k\,\Omega^{J}(u,Iv)\\

&=&G(u,v)+i\,\Omega^{I}(u,v)+j\,\Omega^{K}(Iv,u)-k\,\Omega^{J}(Iu,v)\\

&=&G(u,v)+i\,\Omega^{I}(u,v)+j\,G(KIv,u)-k\,G(JIv,u)\\

&=&G(u,v)+i\,\Omega^{I}(u,v)+j\,G(Jv,u)+k\,G(Kv,u)\\

&=&G(u,v)+i\,\Omega^{I}(u,v)-j\,\Omega^{J}(u,v)-k\,\Omega^{K}(u,v)\\

&\neq& \langle u, v \rangle

\end{eqnarray*}

Did I make something wrong in my elaboration? Can You spot it?

Last edited:

- #6

- 4

- 0

Thanks Jim McNamara

I focus only on a quaternionic vector space case which can be seen as a (linear) hyperkahler manifold.

Share:

- Replies
- 6

- Views
- 7K