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I am familiar with the concept of a dual space $V^*$ as the set of all linear functionals $\tilde{\omega}: V \rightarrow \mathbb{R}$. The inner product on $V$ is usually used to define the dual of a vector $\vec{v}\in V$, that is the dual vector $\tilde{v}$ is the unique linear functional which satisfies \begin{equation} \tilde{v}(\vec{w}) := \langle \vec{v}, \vec{w}\rangle, \end{equation} and the dual basis satisfies $\tilde{e}^\alpha(\vec{e}_\beta) = \delta^\alpha_\beta $.

I am used to thinking about the dual space as a distinct object to the original vector space - that vectors in $V$ cannot be expressed as linear combinations of dual vectors. However, in their book General Relativity: An Introduction for Physicists, the authors Hobson, Efstathiou and Lasenby write that any vector may be written as either a sum of basis vectors or as basis covectors:

$$\vec{v} = v^{\alpha}\vec{e}_\alpha = v_\alpha\vec{e}^\alpha $$

Is this reasonable?

If it is not reasonable, then suppose for some basis $\{\vec{e}_\alpha\}$ of a $3D$ vector space, a set of reciprocal vectors $\{\vec{e}^\alpha\}$ are defined as $$ \vec{e}^1= \frac{\vec{e}_2\times \vec{e}_3}{\vec{e}_1\cdot( \vec{e}_2\times\vec{e}_3)} $$ etc. (this can easily be extended to higher dimensions). These vectors satisfy $\langle{\vec{e}^\alpha}, \vec{e}_\beta\rangle = \delta^\alpha_\beta$, so act as dual vectors, but are still members of $V$. Is there any problem with the quoted equation if these reciprocal vectors are used? Are reciprocal vectors covectors?

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    $\begingroup$ Comments to the post (v1): 1. Presumably Hobson et. al. assume a metric tensor. 2. Btw the cross-product also depends on a metric tensor. 3. Consider to include the extension to higher dimensions. $\endgroup$ – Qmechanic Oct 21 '19 at 10:14
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    $\begingroup$ does not the cross product of two vectors produces pseudovector instead of vector? $\endgroup$ – Umaxo Oct 21 '19 at 12:28
  • $\begingroup$ The notation I was using was just the triple product - I've updated it now. $\endgroup$ – Arthur Morris Oct 21 '19 at 16:46
  • $\begingroup$ The cross-product as such is only defined in 3D space. It's not clear how you want this to generalized to spaces of higher (or lower) dimension. $\endgroup$ – Brick Oct 21 '19 at 17:06
  • $\begingroup$ You can define an n dimensional cross product as an operator which takes in n-1 vectors and spits out a vector which is perpendicular to all of them. Practically, this can be achieved using the determinant. $\endgroup$ – Arthur Morris Oct 21 '19 at 17:40
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For the first part: no, it's not reasonable: any vector $\vec{v}\in V$ can be written $v^\alpha\vec{e}_\alpha$ for some basis $\left\{\vec{e}_\alpha\right\}$, and any vector $\tilde{u} \in V^*$ can be written $\tilde{u}= u_\alpha \tilde{e}^\alpha$, but these are not elements of the same vector space: you can't add them or equate them.

Note that the dual space exists whether or not there is an inner product: it's just the space of linear functions $V \to \mathbb{R}$. And in particular given some basis $\left\{\vec{e}_\alpha\right\}$ for $V$ there is a uniquely-defined dual basis for $V^*$, $\left\{\tilde{e}^\alpha\right\}$, defined by $\tilde{e}^\alpha(\vec{e}_\beta) = \delta^\alpha{}_\beta$: none of this depends on an inner product.

What an inner product gives you is a 1-1 map between vectors and covectors. Given an inner product $\langle\_,\_\rangle$, then you can define $\tilde{v} = \langle\vec{v},\_\rangle$. This in particular gives you a 1-1 relationship between the elements of a basis on $V$ and one on $V^*$: as opposed to their being merely uniquely-defined dual basis without an inner product you can now identify elements of the two bases with each other.

And of course this 1-1 mapping causes people to get lazy and say that $\vec{v}$ is the same as $\tilde{v} = \langle\vec{v},\_\rangle$ but it's not.

In the second part, I think that what you're doing is just a special case of a slightly clever change of basis, and it's easier to understand the general case (for me anyway) because it avoids all the cross-product stuff.

Given two bases $\left\{\vec{e}_\alpha\right\}$ and $\left\{\vec{e}'_\alpha\right\}$, then the relation between them is

$$\vec{e}'_\alpha = \Lambda_\alpha{}^\beta\vec{e}_\beta$$

The corresponding rule for the indices of a vector $\vec{v}$ is

$$v'^\alpha = v^\beta\left(\Lambda^{-1}\right)_\beta{}^\alpha$$

But, given some inner product we can express as a tensor in the usual way, and use it in the usual way to find the components of $\tilde{v}$ in the dual basis:

$$v_\alpha = v^\beta g_{\beta\alpha}$$

Oh, but now we can do a disgusting trick: choose a new basis such that

$$\left(\Lambda^{-1}\right)_\beta{}^\alpha = g_{\beta\alpha}$$

This is fine although it looks horrible: the metric must be nonsingular, so I can simply pick its components as the components of the change-of-basis matrix $\Lambda$. In particular remember that $\Lambda$ isn't a tensor.

And now I get:

$$ \begin{align} v'^\alpha &= v^\beta\left(\Lambda^{-1}\right)_\beta{}^\alpha\\ &= v^\beta g_{\beta\alpha}&&\quad\text{yes, this is OK}\\ &= v_\alpha&&\quad\text{as is this} \end{align} $$

Well, this is just an artifact of picking a suitable change of basis, and it's sort-of the same thing as noticing that $g^{\alpha\beta} = \left(g^{-1}\right)_{\alpha\beta}$ as matrices.

But what it does not mean is that the basis I constructed above is a basis for covectors. I think it's just an interesting property of the metric: perhaps someone who has thought about this harder can say more.

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  • $\begingroup$ The notation was just shorthand for the triple product - I've updated it now. $\endgroup$ – Arthur Morris Oct 21 '19 at 16:47
  • $\begingroup$ My confusion in the last part is that $\vec{v}$ can be written in terms of the reciprocal vectors I defined, and its components in this basis are its covariant components. This would mean that the equation $\vec{v}=v_\alpha\vec{e}^\alpha$ would be valid in this basis. So is there anything wrong with writing that if it's clear that the reciprocal vectors are used and not the dual vectors? $\endgroup$ – Arthur Morris Oct 21 '19 at 16:53

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