The following are from O'Wells' book p.156-157.

Let $E$ be a complex vector space of complex dimension $n$. Let $E'$ be the real dual space to the underlying real vector space of $E$, and let $F = E'\otimes_R\mathbb{C}$ be the complex vector space of complex-valued real-linear mappings of $E$ to $\mathbb{C}$.

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My question is why $h=\sum h_{ij}z_i\otimes\bar{z_j}$? It's not clear why there are no terms like $z_i\otimes z_j$, $\bar{z_i}\otimes z_j$...

Also, I wonder why $S$ is an Euclidean inner product, because if so, if we represent $z_i=x_i+y_i$, $\bar{z_i}=x_i-y_i$, then $S$ would be in the form of $x_i\otimes x_j$+$y_i\otimes y_j$+$x_j\otimes x_i+y_j\otimes y_i$, but the practical computation tells me that $S$ has terms like $x_i\otimes y_j$ with coefficients from $h_{ij}$.


2 Answers 2


The matrix representation of \begin{equation} \underset{\mu\nu}\sum h_{\mu\nu}z_\mu\otimes \bar z_\nu \end{equation} in the case where $\mu=1,2$ and $\nu=1,2$ w.r.t. to the basis $\{z_1, z_2\}$ is \begin{equation} [h]_z=\begin{bmatrix} h_{11}&h_{12}\\ h_{21}&h_{22} \end{bmatrix} \end{equation} that for its hermiticity and putting $h':=\Re h_{\mu\nu}$ and $h'':=\Im h_{\mu\nu}$ it is equal to: \begin{equation} \begin{bmatrix} h'_{11}+ih''_{11}&h'_{12}+ih''_{12}\\ h'_{21}+ih''_{21}&h'_{22}+ih''_{22} \end{bmatrix} = \begin{bmatrix} h'_{11}&h'_{12}+ih''_{12}\\ h'_{12}-ih''_{12}&h'_{22} \end{bmatrix} =\\= \begin{bmatrix} h'_{11}&h'_{12}\\ h'_{12}&h'_{22} \end{bmatrix} +i \begin{bmatrix} 0&h''_{12}\\ -h''_{12}&0 \end{bmatrix} :=[S]_z+[A]_z \end{equation} where $[S]_z$ symmetric and $[A]_z$ antisymmetric are the matrix representation of the real and immaginary parts of $h$ in the given basis.

W.r.t. to the basis $(x_1,x_2,y_1,y_2)$, being \begin{equation} \underset{\mu\nu}\sum h_{\mu\nu}z_\mu\otimes \bar z_\nu=\\ \underset{\mu\nu}\sum h'_{\mu\nu}(x_\mu\otimes x_\nu + y_\mu\otimes y_\nu - ix_\mu\otimes y_\nu +i y_\mu\otimes x_\nu)+h''_{\mu\nu}(x_\mu\otimes y_\nu - y_\mu\otimes x_\nu + ix_\mu\otimes x_\nu +i y_\mu\otimes y_\nu) \end{equation} it is: \begin{equation} [h]_{xy}=\begin{bmatrix} h'_{11}+ih''_{11}&h'_{12}+ih''_{12}&-ih'_{11}+h''_{11}&-ih'_{12}+h''_{12}\\ h'_{21}+ih''_{21}&h'_{22}+ih''_{22}&-ih'_{21}+h''_{21}&-ih'_{22}+h''_{22}\\ ih'_{11}-h''_{11}&ih'_{12}-h''_{12}&h'_{11}+ih''_{11}&h'_{12}+ih''_{12}\\ ih'_{21}-h''_{21}&ih'_{22}-h''_{22}&h'_{21}+ih''_{21}&h'_{22}+ih''_{22}\\ \end{bmatrix} =\\\overset{\begin{matrix}h''_{11}=h''_{22}=0\\h'_{21}=-h'_{12}\end{matrix}}= \begin{bmatrix} h'_{11}&h'_{12}+ih''_{12}&-ih'_{11}&-ih'_{12}+h''_{12}\\ h'_{12}-ih''_{12}&h'_{22}&-ih'_{12}-h''_{12}&-ih'_{22}\\ ih'_{11}&ih'_{12}-h''_{12}&h'_{11}&h'_{12}+ih''_{12}\\ ih'_{12}+h''_{12}&ih'_{22}&h'_{12}-ih''_{12}&h'_{22}\\ \end{bmatrix} =\\= \begin{bmatrix} h'_{11}&h'_{12}&0&h''_{12}\\ h'_{12}&h'_{22}&-h''_{12}&0\\ 0&-h''_{12}&h'_{11}&h'_{12}\\ h''_{12}&0&h'_{12}&h'_{22}\\ \end{bmatrix} +i \begin{bmatrix} 0&h''_{12}&-h'_{11}&-h'_{12}\\ -h''_{12}&0&-h'_{12}&-h'_{22}\\ h'_{11}&h'_{12}&0&h''_{12}\\ h'_{12}&h'_{22}&-h''_{12}&0\\ \end{bmatrix} :=\\:=[S]_{xy}+i[A]_{xy} \end{equation} where $[S]_{xy}$ symmetric and $[A]_{xy}$ antisymmetric are the matrix representation of the real and immaginary parts of $h$ in the given basis.

Generalization to higher dimension is straightforward.


There are in $S$ terms like $x_\mu\otimes y_\nu$, but they appear with the corresponding terms $y_\nu\otimes x_\mu$, both multiplied by the same coefficient.

In particular this terms $x_1\otimes y_1$, $x_2\otimes y_2$, $y_1\otimes x_1$ and $y_2\otimes x_2$ are missing in $[S]_{xy}$ (indeed their coefficients are zero), and the term $x_1\otimes y_2$ appears in $[S]_{xy}$ together with $y_2\otimes x_1$ multiplied by the same coefficient $h''_{12}$ and the $x_2\otimes y_1$ appears in $[S]_{xy}$ together with $y_1\otimes x_2$ multiplied by the same coefficient $-h''_{12}$: so $S$ is symmetric.

Moreover $S$ is positive definite (which together with its symmetry means that $S$ is Euclidean) because $h$ (and, of course, $[h]_{xy}$) is so by definition, and $A(v,v)=0$ as you can see from $[A]_{xy}$.

  • $\begingroup$ Thanks for the answe, but this shows what I said, it does have $x_n\otimes y_m$ term. Moreover by this matrix how to see S is a Euclidean metric? $\endgroup$
    – Danny
    Commented Aug 18, 2018 at 19:34
  • $\begingroup$ you're welcome. Please look at the end of the answer where I added a reply to your comment. $\endgroup$
    – trying
    Commented Aug 18, 2018 at 23:29

$h$ is defined as $$h=\underset{\mu,\nu}\sum h_{\mu\nu}z_\mu\otimes \bar z_\nu $$ that is, with no terms like $z_\mu\otimes z_\nu$, $\bar z_\mu\otimes z_\nu$, $\bar z_\mu\otimes \bar z_\nu$ because otherwise it would not be sesquilinear for the following would not hold: $$h(\xi, \lambda\eta)=\underset{\mu,\nu}\sum h_{\mu\nu}\xi_\mu \overline{\lambda\eta_\nu} =\bar\lambda\underset{\mu,\nu}\sum h_{\mu\nu}\xi_\mu \overline{\eta_\nu} =\bar\lambda h(\xi, \eta)$$

As to your concern on $S$, indeed $S$ contains only terms of the type $x_\mu \otimes x_\nu$, $x_\nu \otimes x_\mu$, $y_\mu\otimes y_\nu$, $y_\nu\otimes y_\mu$:

Being \begin{equation} z_\mu\otimes \bar z_\nu=\frac{z_\mu\otimes \bar z_\nu + \bar z_\nu\otimes z_\mu}{2} + \frac{z_\mu\otimes \bar z_\nu - \bar z_\nu\otimes z_\mu}{2} = \\= z_\mu \bar z_\nu+\frac{1}{2}z_\mu\wedge \bar z_\nu \end{equation}

where I have used a new symmetric product $z_1 z_2=z_2 z_1:=(z_1\otimes z_2 + z_2\otimes z_1)/2$ \begin{equation} h=\underset{\mu\nu}\sum h_{\mu\nu}z_\mu\otimes\bar z_\nu=\underset{\mu}\sum h_{\mu\mu}z_\mu \bar z_\mu +\underset{\mu<\nu}\sum (h_{\mu\nu}+h_{\nu\mu})z_\mu\bar z_\nu +\underset{\mu<\nu}\sum \frac{(h_{\mu\nu}-h_{\nu\mu})}{2}z_\mu\wedge \bar z_\nu \end{equation} but $h$ is Hermitian symmetric, that is, $h_{\mu\nu}=\bar h_{\nu\mu}$ and then $$\frac{h_{\mu\nu}+h_{\nu\mu}}{2}=\Re h_{\mu\nu}:=S_{\mu\nu}$$ $$\frac{h_{\mu\nu}-h_{\nu\mu}}{2}=i\Im h_{\mu\nu}:=iA_{\mu\nu}$$ $$h_{\mu\mu}=\Re h_{\mu\mu}$$ and so \begin{equation} h=\underset{\mu}\sum S_{\mu\mu}z_\mu \bar z_\mu +2\underset{\mu<\nu}\sum S_{\mu\nu}z_\mu\bar z_\nu +j\underset{\mu<\nu}\sum A_{\mu\nu}z_\mu\wedge \bar z_\nu=\\ =\underset{\mu,\nu}\sum S_{\mu\nu}z_\mu\bar z_\nu +\frac{j}{2}\underset{\mu,\nu}\sum A_{\mu\nu}z_\mu\wedge \bar z_\nu \end{equation}

Now being $z_\mu = x_\mu+jy_\mu$ and $z_\nu = x_\nu+jy_\nu$, it follows: \begin{align} z_\mu\bar z_\nu&=x_\mu x_\nu+y_\mu y_\nu-j(x_\mu y_\nu-y_\mu x_\nu)\\ z_\mu\wedge\bar z_\nu&=x_\mu\wedge x_\nu+y_\mu\wedge y_\nu-j(x_\mu\wedge y_\nu-y_\mu\wedge x_\nu) \end{align}

Then $$S=\Re h=\underset{\mu,\nu}\sum S_{\mu\nu}\Re \{z_\mu\bar z_\nu\} -\frac{1}{2}\underset{\mu,\nu}\sum A_{\mu\nu}\Im \{z_\mu\wedge \bar z_\nu\}=\\ =\underset{\mu,\nu}\sum S_{\mu\nu}(x_\mu x_\nu+y_\mu y_\nu) =\frac{1}{2}\underset{\mu,\nu}\sum S_{\mu\nu}(x_\mu \otimes x_\nu+x_\nu \otimes x_\mu+y_\mu\otimes y_\nu+y_\nu\otimes y_\mu)$$ that is, the second term vanishes out because $$\Im \{z_\mu\wedge \bar z_\nu\} = \Im \{z_\nu\wedge \bar z_\mu\}$$

and $$A_{\mu\nu} = -A_{\nu\mu}$$

  • $\begingroup$ I still feel confused why $z_i\otimes\bar{z_j}(m,\lambda n)=m_i\bar{\lambda}\bar{n_j}$? $\endgroup$
    – Danny
    Commented Aug 16, 2018 at 23:57
  • $\begingroup$ And if $h_{mn}=a+ib$, then $h_{mn}z_m\otimes\bar{z_n}=h_{mn}(x_m+iy_m)\otimes(x_n-iy_n)=(a+ib)(x_m\otimes x_n+y_m\otimes y_n+iy_m\otimes x_n+ix_m\otimes y_n)$, then there is a term $-bx_m\otimes y_n$, which can not vanish out. So there is a contradiction? $\endgroup$
    – Danny
    Commented Aug 17, 2018 at 1:21
  • $\begingroup$ Moreover, if it's an Euclidean inner product, shouldn't it be in the form of $\sum x_m\otimes x_m+y_m\otimes y_m$? $\endgroup$
    – Danny
    Commented Aug 17, 2018 at 3:00
  • $\begingroup$ $z_i\otimes z_j(m, \lambda n)\neq m_i\bar\lambda\bar n_i$, because is not Hermitian, while $h$ is, because the matrix $(h_{\mu\nu})$ is Hermitian symmetric. Moreover, I don't see contraddictions: terms like $x_m\otimes y_n$ vanishes out again because of $h_{\mu\nu}$ Hermtian symmetry; it does not vanish for a generic $h_{\mu\nu}=a+jb$. Thirdly, an inner product is Euclidean not only when it has the form $\sum x_m\otimes x_n+y_m\otimes y_m$ (see inner product on Wikipedia), but when there is a coordinate system in which the inner product can be expressed by such a form and $S$ is of this kind. $\endgroup$
    – trying
    Commented Aug 17, 2018 at 6:05
  • $\begingroup$ By Hermitian symmetry, I think you are considering using the terms from $h_{nm}z_n\otimes\bar{z_m}=(a-b_i)(x_n\otimes x_m+y_n\otimes y_m+iy_n\otimes x_m+ix_n\otimes y_m$ to kill the term $x_m\otimes y_n$, but unfortunately, there is no such a term. $\endgroup$
    – Danny
    Commented Aug 17, 2018 at 6:35

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