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I am currently taking a module on Differential Geometry, and as such I have been mostly looking at tensors as objects following the usual transformation rule. However I would like to understand how that is equivalent to viewing tensors as multilinear maps from direct products of Vector and their Dual spaces to the corresponding field of scalars.

Here is my understanding of the distinction between a covector and a vector(as multilinear maps):

Let $V$ be a vector space over $\mathbb{R}$. The dual space of $V$, denoted $V^*$, is defined as $V^* =\{ϕ:V→\mathbb{R}∣ϕ$ is linear$\}$. $V^*$ is an n-dimensional vector space over the same field. Thus, $V$ is isomorphic to $V^*$. Thus they are, as vector spaces, identical. But we consider them to be different by considering elements of $V^*$(covectors) to act on vectors as linear maps.

My confusion:

I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?

My attempted explanation for this:

Either $V$ and $V^*$ can be considered the dual space of the other, the choice is arbitrary

It seems to me that when viewing vectors and covectors as objects obeying their respective transformation rule, this confusion is not an issue, since we simply define the operation of contraction to combine a vector and a covector into a scalar.

I really can't make intuitive sense of this, so any insights would be appreciated.

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    $\begingroup$ You are exactly right. The choice of what is a covector and what is a vector is completely arbitrary. However, there are certain standards that we tend to stick to. $\endgroup$
    – K.defaoite
    Nov 30, 2020 at 23:18
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    $\begingroup$ The claim is not true in full generality. If $V$ is a finite-dimensional vector space then, yes, $V$ is isomorphic to $V^*$. But in general they are different. $\endgroup$
    – Zhen Lin
    Nov 30, 2020 at 23:48

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Thus they are, as vector spaces, identical.

Be careful. It's dangerous to consider mathematical objects to be identical just because there happens to be an isomorphism between them. The problem is that isomorphisms that fail to be "natural" generalize poorly. For example, if $V$ carries a representation of a group $G$, $V^{\ast}$ also carries a representation of $G$, and it will generally be a non-isomorphic representation. There are many other contexts like this, for example in differential geometry, where you'll get very confused if you carelessly identify a vector space with its dual. Tangent vectors are not cotangent vectors and vector fields are not differential $1$-forms.

I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?

This is a point that is very often poorly explained. The resolution is that the term "covector" is relative, and what it's relative to is a choice of "base" vector space to act as the "vectors." That is, the grammar of the term "covector" is "an object is a covector with respect to $V$," which means that it's an element of the dual space $V^{\ast}$.

So it's true that $V^{\ast}$ consists of the covectors with respect to $V$, and it's also true that $V$ consists of the covectors with respect to $V^{\ast}$. Formally what this means is that there is a natural double dual isomorphism $V \to (V^{\ast})^{\ast}$ (when $V$ is finite-dimensional); this time it is actually mostly harmless to consider these as identical, due to naturality. For example the double dual of a $G$-representation is a naturally isomorphic $G$-representation.

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This mainly is a question on linear algebra, and it touches on a subtle point. As you say, two (finite dimensional) vector spaces of the same dimension are isomorphic. But this does not mean that they are (or can be considered as) identical. Indeed, there is a big freedom of choice for the isomorphism between two spaces of the same dimension. In particular, if you look at an abstract vector space $V$, you know that $V^*$ has the same dimension, but there is no "natural" way to associate to $v\in V$ a linear functional in $V^*$. To get such an association you need to make additional choices, e.g. a basis for $V$ or an inner product on $V$. This is related to the situation you meet in differential geometry, where you know that all tangent spaces have the same dimension, but there is no natural way to identify different tangent spaces without making additional choices (say a chart). So you cannot simply consider them as "identical" although they are isomorphic. In the same spirit, you have to distinguish between a tangent space and its dual space.

Still, there is a certain symmetry in the situation between $V$ and $V^*$. This comes from the fact that you can also consider the dual $(V^*)^*$ of $V^*$. Of course, this has again the same dimension as $V^*$ and hence as $V$, but here much more is true: There is a natural way to identify the space $(V^*)^*$ with $V$. (But there is no natural way to identify $(V^*)^*$ with $V^*$.) The natural identification sends a vector $v\in V$ to the linear map $V^*\to\mathbb R$ given by $\phi\mapsto\phi(v)$. This means that in a way you can swap the roles of $V$ and $V^*$ (and this is used frequently when dealing with tensor fields in differential geometry), but this does not mean that you have a way to identify $V$ with $V^*$.

This does not mean that there is no difference between vectors and covectors. From a differential geometry point of view, this is because smooth maps (and diffeomorphisms) act on them differently. Consequently, there is a diffeomorphism invariant linear operation on covectors (the exterior derivative) and no such operation exists on vectors. But you can also interpret vectors in terms of linear functionals.

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  • $\begingroup$ I am not 100% sure I understand why there is no natural way to identify $V$ with $V^*$. Upon trying to think of a "natural" mapping, I realised that every mapping required the specification of a specific basis in $V$ and a specific basis in $V^*$. Thus, I ask: is the reason why there is no "natural" identification of $V$ with $V^*$ because such an identification would rely on a choice of basis for both spaces, and this choice is arbitrary? $\endgroup$ Dec 1, 2020 at 20:45
  • $\begingroup$ Yes, this is correct. $\endgroup$ Dec 2, 2020 at 15:45

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