# Inner product of vectors and covectors

1. Oct 26, 2011

### Wox

I'm struggling with the meaning of inner products between vectors and covectors.

Consider dual space $V^{\ast}$ of an inner product space V over field K, with linear forms $\vec{v}\:^{\ast}\in V^{\ast}$

$\vec{v}\:^{\ast}:V\rightarrow K:\vec{w}\rightarrow\left<\vec{v},\vec{w}\right>$

In Einstein notation we can write vector and covector with respect to basis and dual basis: $\vec{v}=v^{i}\vec{b}\;_{i}\quad$ and $\vec{v}\:^{\ast}=v_{i}\vec{b}\;^{i}\quad$. Because of the inner product, V is isomorphic to its dual $V\cong V^{\ast}$ and we can associate $\vec{v}\:^{\ast}\quad$ with its preimage under this isomorphism, which is $\vec{v}\quad$ after a change of basis. Now we can write following inner products

$\left<\vec{v},\vec{w}\right>=v^{i}w^{j}\left<\vec{b}\;_{i},\vec{b}\;_{j}\right>$

$\left<\vec{v}\:^{*},\vec{w}\right>=v_{i}w^{j}\left<\vec{b}\;^{i},\vec{b}\;_{j}\right>=v_{i}w^{i}$

$\left<\vec{v},\vec{w}\:^{*}\right>=v^{i}w_{j}\left<\vec{b}\;_{i},\vec{b}\;^{j}\right>=v^{i}w_{i}$

$\left<\vec{v}\:^{\ast},\vec{w}\:^{\ast}\right>=v_{i}w_{j}\left<\vec{b}\;^{i},\vec{b}\;^{j}\right>$

Question: what do these inner products mean and what is their relation?

I would say that they are all the same since vector and (preimage of) covector are the same abstract vector but given with respect to a different basis. However when I try this for an example, the evalutation of the inner products gives different results (only the first and last inner product are equal).

I'm also wondering what these inner products mean. The first inner product is obvious. Since $V\cong V^{\ast}$, the second inner product is equivalent to the evaluation of $\vec{v}\:^{\ast}(\vec{w})\quad$ so that $\vec{v}\:^{\ast}(\vec{w})\equiv\left<\vec{v}\:^{*},\vec{w}\right>=\left<\vec{v},\vec{w}\right>$. As for the third and fourth inner product, I have no idea.

Last edited: Oct 26, 2011
2. Oct 26, 2011

### Wox

An example: consider Euclidean vector space with as inner product the dot product. Suppose $\{\vec{e}\;_{i}\}\$ is the canonical basis, $\{\vec{b}\;_{i}\}\$ the basis in which we work and $\{\vec{b}\;^{i}\}\$ the dual basis.

Suppose the change of basis matrix from $\{\vec{e}\;_{i}\}\$ to $\{\vec{b}\;_{i}\}\$ is given by

$C=\left[\begin{matrix} 1 &-\frac{1}{2}&0\\ 0&\frac{\sqrt{3}}{2}&0\\ 0&0&1 \end{matrix}\right]$

where column i the coordinates of vector $\vec{b}\;_{i}\$ with respect to the canonical basis. The metric tensor $M\;_{i,j}=\left<\vec{ b}\;_{i},\vec{b}\;_{j}\right>=\vec{b}\;_{i}\cdot \vec{b}\;_{j}$ of basis $\{\vec{b}\;_{i}\}\$ is given by

$M=C^{T}C=\left[\begin{matrix} 1 &-\frac{1}{2}&0\\ -\frac{1}{2}&1&0\\ 0&0&1 \end{matrix}\right]$

The change of basis matrix from $\{\vec{e}\;_{i}\}\$ to $\{\vec{b}\;^{i}\}\$ is given by

$C^{\ast}=\left[\begin{matrix} \frac{2}{\sqrt{3}} &\frac{1}{\sqrt{3}}&0\\ 0&1&0\\ 0&0&1 \end{matrix}\right]$

where column i the coordinates of vector $\vec{b}\;^{i}\$ with respect to the canonical basis. The metric tensor $M\;^{i,j}=\left<\vec{ b}\;^{i},\vec{b}\;^{j}\right>=\vec{b}\;^{i}\cdot \vec{b}\;^{j}$ of basis $\{\vec{b}\;^{i}\}\$ is given by

$M^{\ast}=C^{*T}C^{\ast}=\left[\begin{matrix} \frac{4}{3} &\frac{2}{3}&0\\ \frac{2}{3}&\frac{4}{3}&0\\ 0&0&1 \end{matrix}\right]$

The change of basis matrix from $\{\vec{b}\;_{i}\}\$ to $\{\vec{b}\;^{i}\}\$ is given by

$D^{\ast}=C^{-1}C^{\ast}=\left[\begin{matrix} \frac{2}{\sqrt{3}} &\frac{2}{\sqrt{3}}&0\\ 0&\frac{2}{\sqrt{3}}&0\\ 0&0&1 \end{matrix}\right]$

where column i the coordinates of vector $\vec{b}\;^{i}\$ with respect to basis $\{\vec{b}\;_{i}\}\$.

Now we can take two vectors and evaluate the four inner products

$\vec{v}=\vec{b}\;_{1}$

$V=\left[\begin{matrix}1\\0\\0\end{matrix}\right]$

$V^{\ast}=D^{\ast\;-1}V=\left[\begin{matrix}\frac{\sqrt{3}}{2}\\0\\0\end{matrix}\right]$

$\vec{w}=\vec{b}\;_{1}+\vec{b}\;_{2}$

$W=\left[\begin{matrix}1\\1\\0\end{matrix}\right]$

$W^{\ast}=D^{\ast\;-1}W=\left[\begin{matrix}0\\\frac{\sqrt{3}}{2}\\0\end{matrix}\right]$

$\left<\vec{v},\vec{w}\right>=V^{T}MW=\frac{1}{2}$

$\left<\vec{v}\:^{*},\vec{w}\right>=V^{\ast\; T}W=\frac{\sqrt{3}}{2}$

$\left<\vec{v},\vec{w}\:^{*}\right>=V^{T}W^{\ast}=0$

$\left<\vec{v}\:^{*},\vec{w}\:^{*}\right>=V^{\ast\;T}M^{*}W^{\ast}=\frac{1}{2}$

Only the first and last are the same, while I would expect all of them beeing the same.

Last edited: Oct 26, 2011
3. Oct 26, 2011

### Deveno

i am confused by your assertion in the first post that:

<bi,bj> = δij.

in fact, i don't see how you have even given a definition of what <bi,bj> even is (the two vectors live in two different vector spaces).

if by <v*,w> you mean <ψ-1(v),w>, where ψ: V→V* is the isomorphism given by:

ψ(v) = <v,_> then i think you should PROVE <bi,bj> = δij, which it seems to me it may not.

4. Oct 27, 2011

### Wox

This is exactly my problem. What do these inner products mean? Note that I didn't invent them myself. They are used for example in http://dx.doi.org/10.1107/97809553602060000549" [Broken] of the International Tables For Crystallography.

As for your question, suppose V has an inner product and $F_{\left<\cdot,\cdot\right>}$ the isomorphism with its dual, induced by the inner product

$F_{\left<\cdot,\cdot\right>}:V\rightarrow V^{\ast}:\vec{v}\rightarrow\left<\vec{v},\cdot \right>$

$\left<\vec{v},\cdot\right> : V\rightarrow K: \vec{w}\rightarrow \left<\vec{v},\vec{w}\right>$

If we define $\left<\vec{v},\cdot\right>(\vec{w})\equiv\vec{v}\:^{\ast}(\vec{w})\equiv\left<\vec{v}\:^{*} ,\vec{w}\right>$ then we can write (by definition)

$\left<\vec{v}\:^{*} ,\vec{w}\right>=\left<\vec{v},\vec{w}\right>=\left<F^{-1}_{\left<\cdot,\cdot\right>}(\vec{v}\:^{*}),\vec{w}\right>$

Lets now evaluate this inner product. Suppose the inner product is the Euclidean one (dot product of vectors). We need to consider a basis in direct and dual space. If $\{\vec{b}\;_{j}\}$ is a basis of V then the set of covectors $\{\vec{b}\;^{i}\}$ for which $\vec{b}\;^{i}(\vec{b}\;_{j})=\delta^{i}_{j}\$ is a basis of dual space, called the dual basis (I think it is only a basis when V is finite, so suppose finite dimension). Using the same notation as above

$\left<\vec{b}\;^{i} ,\vec{b}\;_{j}\right>=\delta^{i}_{j}=\left<F^{-1}_{\left<\cdot,\cdot\right>}(\vec{b}\;^{i}),\vec{b}\;_{j}\right>$

Now we can evaluate the inner product

$\left<\vec{v}\:^{*} ,\vec{w}\right>=\left<F^{-1}_{\left<\cdot,\cdot\right>}(\vec{v}\:^{*}),\vec{w}\right>=\left<\sum_{i}v_{i}F^{-1}_{\left<\cdot,\cdot\right>}(\vec{b}\;^{i}),\sum_{j}w^{j}\vec{b}\;_{j}\right>=\sum_{i,j}v_{i}w^{j} \left<F^{-1}_{\left<\cdot,\cdot\right>}(\vec{b}\;^{i}),\vec{b}\;_{j}\right>=\sum_{i,j}v_{i}w^{j} \delta^{i}_{j}=\sum_{i}v_{i}w^{i}\equiv v_{i}w^{i}$

On the other hand we know that

$\left<\vec{v}\:^{*} ,\vec{w}\right>=\left<\vec{v},\vec{w}\right>=\left<\sum_{i}v^{i}\vec{b}\;_{i},\sum_{j}w^{j}\vec{b}\;_{j}\right>=\sum_{i,j}v^{i}w^{j} \left<\vec{ b}\;_{i},\vec{b}\;_{j}\right>=\sum_{i,j}v^{i}w^{j} M_{i,j}\equiv v^{i}w^{j} M_{i,j}$

This would mean that $v_{i}w^{i}=v^{i}w^{j} M_{i,j}$ which is not true in the example I gave. So either my reasoning is wrong or I made a mistake in the example.

Last edited by a moderator: May 5, 2017
5. Oct 27, 2011

### Wox

The four inner products are equal in the example. There was a mistake in $C^{\ast}$ which should be

$C^{\ast}=\left[\begin{matrix} 1 &0&0\\ \frac{1}{\sqrt{3}}&\frac{2}{\sqrt{3}}&0\\ 0&0&1 \end{matrix}\right]$

As a result, the inner products are equal:

$D^{\ast}=C^{-1}C^{\ast}=\left[\begin{matrix} \frac{4}{3} &\frac{2}{3}&0\\ \frac{2}{3}&\frac{4}{3}&0\\ 0&0&1 \end{matrix}\right]$

$V^{\ast}=D^{\ast\;-1}V=\left[\begin{matrix}1\\ \frac{-1}{2}\\0\end{matrix} \right]$

$W^{\ast}=D^{\ast\;-1}W=\left[\begin{matrix}\frac{1}{2}\\ \frac{1}{2}\\0\end{matrix} \right]$

$\left<\vec{v},\vec{w}\right>=V^{T}MW=\frac{1}{2}$

$\left<\vec{v}\:^{*},\vec{w}\right>=V^{\ast\; T}W=\frac{1}{2}$

$\left<\vec{v},\vec{w}\:^{*}\right>=V^{T}W^{\ast}= \frac{1}{2}$

$\left<\vec{v}\:^{*},\vec{w}\:^{*}\right>=V^{\ast\; T}M^{*}W^{\ast}=\frac{1}{2}$

This confirms my idea that all these inner products must be the same because $\vec{v}\:^{\ast}\equiv F^{-1}_{\left<\cdot,\cdot\right>}(\vec{v}\:^{\ast})$ and $\vec{v}\quad$ are the same abstract vectors but given with respect to a different basis. However I can only explain $\vec{v}\:^{\ast}\equiv F^{-1}_{\left<\cdot,\cdot\right>}(\vec{v}\:^{\ast})$ in the inner product $\left<\vec{v}\:^{*},\vec{w}\right>$ (see previous post) but not in $\left<\vec{v},\vec{w}\:^{*}\right>$ or $\left<\vec{v}\:^{*},\vec{w}\:^{*}\right>$. Any ideas?

6. Oct 27, 2011

### Ben Niehoff

As Deveno mentioned, strictly speaking, it does not make sense to speak of "inner products" between objects from different vector spaces.

Many physicists, however, have adopted the bastard notation $\langle w, v\rangle$ to mean $w(v)$ when $w \in V^*, \; v \in V$.

As for the notation $\langle v, w \rangle$, recall that for finite-dimensional vector spaces, $V \simeq V^{**}$.

If one defines a map $\varphi : V^* \rightarrow V$ such that

$$\langle \varphi(w), v \rangle = w(v)$$

then the following should all give identical results:

$$w(v), \qquad \langle \varphi(w), v \rangle_V, \qquad \langle w, \varphi^{-1}(v) \rangle_{V^*}, \qquad \big(\varphi^{-1}(v)\big) \big( \varphi(w) \big)$$

which your source has (confusingly) chosen to write as

$$\langle w, v \rangle, \qquad \langle w^*, v \rangle, \qquad \langle w, v^* \rangle, \qquad \langle w^*, v^* \rangle$$

7. Oct 28, 2011

### Wox

Ok the "inner product" is just a notation issue but still, I understand that $w(v)=\langle \varphi(w), v \rangle$ because of canonical isomorphism $\varphi : V^* \rightarrow V$. However I do not see why $\langle w, \varphi^{-1}(v) \rangle_{V^*}$ and $\big(\varphi^{-1}(v)\big) \big( \varphi(w) \big)$ are also equal to $w(v)$. You mentioned the isomorphism with the double dual, but I don't see how this helps here...

8. Oct 28, 2011

### Ben Niehoff

The inner product $\langle \cdot, \cdot \rangle_{V^*}$ is defined to be compatible with the canonical isomorphism. That's what makes the isomorphism canonical!

In a basis, one can write for $v_1, v_2 \in V$

$$\langle v_1, v_2 \rangle_V = v_1^\top G v_2$$

where G is some symmetric, positive definite matrix. Then for $w_1, w_2 \in V^*$, we have

$$\langle w_1, w_2 \rangle_{V^*} = w_1 G^{-1} w_2^\top$$

(remember that members of $V^*$ are row vectors). The canonical isomorphism $\varphi : V^* \rightarrow V$ can then be written

$$\varphi(w) = G^{-1} w^\top$$

Then it should be clear (using the symmetry of G) that

$$\langle \varphi(w_1), \varphi(w_2) \rangle_V = \langle w_1, w_2 \rangle_{V^*}$$

9. Oct 29, 2011

### Wox

Ok, I think I understand it now. A non-degenerate bilinear form on a finite vector space induces the following isomorphism with its dual

$F_{\left<\cdot,\cdot\right>}:V\rightarrow V^{\ast}:\vec{v}\rightarrow\left<\vec{v},\cdot \right>$

$\left<\vec{v},\cdot \right> : V\rightarrow K: \vec{w}\rightarrow \left<\vec{v},\vec{w}\right>$

In different notation this is $\left<\vec{v},\cdot \right>(\vec{w})\equiv \left<\vec{v}\:^{*},\vec{w}\right> =\left<\vec{v},\vec{w}\right>$. If we now define a mapping

$f:V^{\ast}\times V^{\ast}\rightarrow K: (\vec{v}\:^{\ast},\vec{w}\:^{\ast})\rightarrow \left<F_{\left<\cdot,\cdot\right>}^{-1}(\vec{v}),F_{\left<\cdot,\cdot\right>}^{-1}(\vec{w})\right>=\left<\vec{v},\vec{w}\right>$

then it can be seen that it is linear in both arguments and hence a bilinear form on $V^{\ast}$ which we can write in different notation $f(\vec{v}\:^{\ast},\vec{w}\:^{\ast})\equiv \left<\vec{v}\:^{*},\vec{w}\:^{*}\right>=\left< \vec{v},\vec{w}\right>$

So $\left<\vec{v}\:^{*},\vec{w}\right> = \left<\vec{v},\vec{w}\right>$ determines the isomorphism $F_{\left<\cdot,\cdot\right>}:V\rightarrow V^{\ast}$ and $\left<\vec{v}\:^{*},\vec{w}\:^{*}\right> = \left<\vec{v},\vec{w}\right>$ is an bilinear form induced by the isomorphism.

So we only have to define what $\left<\vec{v},\vec{w}\:^{*}\right>$ might mean in the paper I referred to. Can't we find a second isomorphism between $V$ and $V^{\ast}$

$\varphi_{\left<\cdot,\cdot\right>}:V\rightarrow V^{\ast}: \vec{v}\rightarrow\left<\cdot,\vec{v}\right>$

$\left<\cdot,\vec{v}\right> : V\rightarrow K: \vec{w}\rightarrow \left<\vec{w},\vec{v}\right>$

In different notation this is $\left<\cdot,\vec{v} \right>(\vec{w})\equiv \left<\vec{w},\vec{v}\:^{*}\right> =\left<\vec{w},\vec{v}\right>$. So $\left<\vec{v},\vec{w}\:^{*}\right> = \left<\vec{v},\vec{w}\right>$ determines the isomorphism $\varphi_{\left<\cdot,\cdot\right>}:V\rightarrow V^{\ast}$

Last edited: Oct 29, 2011