Are vectors and covectors the same thing?

Question

In Euclidean space, we usually don't distinguish between vectors and covectors (or dual vectors or 1-forms or whatever you want to call them) -- because the spaces overlap. However, a physicist friend of mine (I'm a physics major, too, BTW) was trying to convince me that not only are Euclidean vectors and covectors the same thing, but ANY vector is equivalent to a covector. His reasoning is that because there is an isomorphism between them and we can convert one to the other with the metric tensor, that there's no real difference between them. My argument is that an isomorphism between mathematical objects is not enough to call two things the same. For example, there is an isomorphism between the additive groups of $\Bbb R$ and $\Bbb R^2$, but surely no one would say that $\Bbb R$ is the SAME THING as $\Bbb R^2$.

So which of us is right? Are vectors the same thing as covectors?

Answer 1

To add to all the answers above, there is a delightful example in the text "Mathematics for Physics" by Stone and Goldbart, (Appendix A.3), to clarify the difference between vectors and co-vectors, which I can't resist quoting here.

One way of driving home the distinction between $V$ and $V^*$ is to consider the space $V$ of fruit orders at a grocers. Assume that the grocer stocks only apples, oranges and pears. The elements of $V$ are then vectors such as

$x = 3 \text{ kg apples }+ 4.5 \text{ kg oranges } + 2 \text{ kg pears.} $

Take $V^*$ to be the space of possible price lists, an example element being

$f = (\$3.00/\text{kg}) \text{ apples}^* + (\$2.00/\text{kg}) \text{ oranges}^* + (\$1.50/\text{kg}) \text{ pears}^*$

The evaluation of $f$ on $x$

$f(x) = 3 \times \$3.00 + 4.5 \times \$2.00 + 2 \times \$1.50 = 21.0$

then returns the total cost of the order. You should have no difficulty in distinguishing between a price list and box of fruit!

Answer 2

If $V$ is a finite-dimensional vector space then there exists an isomorphism $V\cong V^*$, under which every vector in $V$ can be interpreted as a covector (linear map $V\to $ the scalar field). However, there are going to be many such maps, and there is in general no canonical such isomorphism, so given a vector $v\in V$, which covector in $V^*$ it corresponds to depends on an ultimately arbitrary choice of isomorphism $V\cong V^*$.

Your friend does observe a valid mental tool that is useful in mathematics: identifying things when they are "essentially the same." If ever there is a canonical isomorphism $A\cong B$ between two objects satisfying some set of desirable properties (what's desirable depends on context), then we can think of any $a\in A$ as a corresponding $b\in B$. (Often we're really speaking about a collection of objects $A$ that are built or result from some kind of process or characterization, and we may use the language of category theory (universal properties, natural isomorphisms). In particular, there does not exist a natural isomorphism between the identity endofunctor and "dual space" endofunctor on the category of finite-dimensional vector spaces over a given field.)

But this identification idea doesn't work when there is no unique or canonical isomorphism between two given objects or structures. Indeed, $\{1,2,3\}\cong\{\rm\color{Red}{red},\color{Green}{green},\color{Blue}{blue}\}$. But if someone were to claim that "the numbers $1,2,3$ are basically colors," you could simply ask "if $1$ is a color, which color is it specifically?" and there would be no correct answer. Whenever there is a unique or canonical isomorphism $A\cong B$, given a $a\in A$, the question "which $b\in B$ is $a$ basically?" does have a unique correct answer; it's $\phi(a)$, where $\phi:A\to B$ is the aforementioned isomorphism.

Answer 3

Expanding just a touch on my comment:

One simple way to see that vectors and covectors are not as naively identifiable as your friend says is to realize that there are settings where the spaces they form aren't naturally isomorphic. They are developed independently on abstract manifolds, via the tangent and cotangent structures, respectively.

In particular, the only reason they are naturally isomorphic in Euclidean space is the presence of a natural Riemannian metric. It's this metric that gives you the sort structure you need to convert covectors (actually covector fields) into vectors (actually vector fields). This mapping is called the "musical isomorphism." You can get a similar mapping any time you have a non-degenerate two-form, e.g. a symplectic form would also work. But in the absence of such structure -- on an abstract finite-dimensional smooth manifold -- there is no such natural isomorphism, and you can't naturally identify the tangent and the cotangent structures (although any given tangent and cotangent space are isomorphic, by a dimensional argument).

This is a really important distinction, even in physics, as many have pointed out. It will show up when you learn how to integrate differential forms; you can't talk about "gradients" unless you have a Riemannian metric. It will also show up when you learn the symplectic structure of Hamiltonian mechanics: the natural setting for that is the cotangent bundle, but you can use the Legendre transform, sort of like the musical isomorphism, to push that physics into the tangent bundle, which gives you Lagrangian mechanics.

EDIT

You might also be interested in the following discussion of how to "visualize" a covector, which can help to give an intuitive sense of why, though isomorphic, covectors and vectors are different things:

http://www.physicsinsights.org/pbp_one_forms.html