A Is the Inner Product in Quaternionic Vector Spaces Truly Hyperhermitian?

[Leditto]

Let $V$ be a quaternionic vector space with quaternionic structure $\{I,J,K\}$. One can define a Riemannian metric $G$ and hyperkahler structure $\{\Omega^{I},\Omega^{J}, \Omega^{K}\}$. Do this inner product
$$\langle p,q \rangle := G(p,q)+i\Omega^{I}(p,q)+j\Omega^{J}(p,q)+k\Omega^{K}(p,q)$$
really satisfy hyperhermitian condition?

 

[chiro]

Hey Leditto.

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?

 

[Ben Niehoff]

Hey Leditto.

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.

 

[jim mcnamara]

Hmm. maybe this might help:

http://arxiv.org/abs/math/0105206

 

[Leditto]

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?

 

[Leditto]

Hmm. maybe this might help:

http://arxiv.org/abs/math/0105206

Thanks Jim McNamara

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