I'm working through Jeevanjee's "An Introduction to Tensors and Group Theory for Physicists", and while trying to prove that the matrix representation $M(A^\dagger)$ of a Hermitian adjoint $A^\dagger$ is the conjugate-transpose of the matrix $M(A)$, I cannot figure out how to move past the gap between dual vectors and so-called 'metric dual' vectors.

Given an orthonormal basis $\{e_i\}_{i=1,\ldots,n}$ of vector space $V$, he defines a dual basis $\{e^i\}_{i=1,\ldots,n} \in V^*$ such that $e^i(e_j)=\delta_{ij}$.

He then defines the metric dual of a vector $v \in V$ as $\tilde{v} \in V^*$ with $\tilde{v}(w) = (v|w)$, where $(\cdot|\cdot)$ is a non-degenerate Hermitian form and $w \in V$.

In proofs of $M(A^\dagger)=\overline{M(A)}^T$, it seems necessary to use the fact that $(\cdot|\cdot)$ is Hermitian (i.e. $(v|w)=\overline{(w|v)}$), but I don't see how to move from $M(A^\dagger)_j^i = e^i(A^\dagger e_j)$ to $M(A^\dagger)_j^i \stackrel{?}{=} \tilde{e_i}(A^\dagger e_j) = (e_i| A^\dagger e_j)$ in the general case where it the metric is not necessarily Euclidean. That is to say, $(e_i|e_j)$ is not necessarily $\delta_{ij}$, and equivalently $\tilde{e_i}$ is not necessarily $e^i$.

Of course, since the map $e_i \mapsto \tilde{e_i}$ is bijective, $\{\tilde{e_i}\}_{i=1,\ldots,n}$ is a basis of $V^*$, and we can write $e^i = \sum_k a_{ik} \tilde{e_k}$ for scalars $a_{ik}$, giving

$$\begin{align} M(A^\dagger)_j^i &= e^i(A^\dagger e_j) \\ &= \sum_k a_{ik} \tilde{e_k} (A^\dagger e_j) \\ &= \sum_k a_{ik} (e_k | A^\dagger e_j) \\ &= \sum_k a_{ik} (A^\dagger e_j | e_k )^* \\ &= \sum_k a_{ik} (e_j | A e_k )^*. \end{align}$$

Then taking the dual of the dual to be itself, $\sum_k a_{ik} e_k = \tilde{e^i}$ and $$M(A^\dagger)_j^i = (e_j|A\tilde{e^i})^*.$$

But is not $\tilde{e^i} = e_i$ true only for the case of the Euclidean metric where $(e_i|e_j) = \delta_{ij}$?


You're absolutely right. You can only identify the dual space with the original Hilbert space when you can. (Tautology!) Which is almost never. Don't do that for the general proof. You need to think of the dual vectors as living somewhere else entirely and not at all related to the regular vectors except as you've said in your definition. Even in non-Euclidean spaces it is still true that the dual space basis may be defined to give $e^i(e_j) = \delta^i_j$

There are many proofs of this theorem. For the basic properties see:


or google it.

  • $\begingroup$ Sorry, I just realised it was stated in a previous part of the problem that the Hermitian form being considered is positive-definite, so that $(e_i|e_j)=\delta_{ij}$ and the conjugate-transpose holds for this case. Thanks for your help. $\endgroup$ – user37160 Aug 16 '14 at 0:21

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