# What is the difference between a dual vector and a reciprocal vector?

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?

• 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. Oct 21, 2019 at 10:14
• does not the cross product of two vectors produces pseudovector instead of vector? Oct 21, 2019 at 12:28
• The notation I was using was just the triple product - I've updated it now. Oct 21, 2019 at 16:46
• 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. Oct 21, 2019 at 17:06
• 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. Oct 21, 2019 at 17:40

With respect to your question "Are reciprocal vectors covectors" I asked the same question about 6 months ago on a number of forums. No one could give me an answer. I took some time but finaly conviced my self that the answer is yes but had no second proof backing me up. I have just found a paper at the link;

https://www.iucr.org/__data/assets/pdf_file/0017/13193/4.pdf

Which says YES Reciprocal vectors are the covectors of the Real space. See Section 3. Of course I cannot vouch for the credentials of the Author so will leave that with you.

Hope this helps.

Paul

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.

• The notation was just shorthand for the triple product - I've updated it now. Oct 21, 2019 at 16:47
• 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? Oct 21, 2019 at 16:53