The Unapologetic Mathematician

Euclidean Spaces

In light of our discussion of differentials, I want to make a point here that is usually glossed over in most treatments of multivariable calculus. In a very real sense, the sources and targets of our functions are not the vector spaces $\mathbb{R}^n$.

Let’s think about what we need to have a vector space. We need a way to add vectors and to multiply them by scalars. Geometrically, addition proceeds by placing vectors as arrows “tip-to-tail” and filling in the third side of the triangle. Scalar multiplication takes a vector as an arrow and stretches, shrinks, or reverses it depending on the value of the scalar. But both of these require us to think of a vector as an arrow which points from the origin to the point with coordinates given by the components of our vector.

But this makes the origin a very special point indeed. And why should we have any such special point, from a geometric perspective? We already insisted that we didn’t want to choose a basis for our space that would make some directions more special than others, so why should we have to choose a special point?

What really matters in our spaces is their topology. But we don’t want to forget all of the algebraic structure either. There are still some vestiges of the structure of a vector space that still make sense in the absence of an origin. Indeed, we can still talk about it as an affine space, where the idea of displacement vectors between points still makes sense. And these displacement vectors will be actual vectors in $\mathbb{R}^n$. Like any torsor, this means that our space “looks like” the group (here, vector space) we use to describe displacements, but we’ve “forgotten” which point was the origin. We call the result a “Euclidean” space, since such spaces provide nice models of the axioms of Euclidean geometry.

So let’s try to be a little explicit here: we actually have two different kinds of geometric objects floating around right now. First are the points in an $n$-dimensional Euclidean space. We can’t add these points, or multiply them by scalars, but we can find a displacement vector between two of them. Such a displacement vector will be in the $n$-dimensional real vector space $\mathbb{R}^n$. When it’s convenient to speak in terms of coordinates, we first pick an (arbitrary) origin point. Now if we’re sloppy we can identify a point in the Euclidean space with its displacement vector from the origin, and thus confound the Euclidean space of points and the vector space of displacements. We can proceed to choose a basis of our vector space of displacements, which gives coordinates to the Euclidean space of points; the point $(x^1,\dots,x^n)$ is the one whose displacement vector from the origin is $x^ie_i$.

Now, the rant. Some multivariable calculus books are careful about not doing nonsense things like “adding” or “scalar multiplying” points, but many do exactly these sorts of things, giving the impression to students that points are vectors. Even among the texts that are careful, I don’t recall seeing any that actually go so far to mention that a point is not a vector. When I teach the course I’m careful to point out that they’re not quite the same thing (though not in quite as much detail as this) and I go so far as to write them differently, with vector coordinates written out between angle brackets instead of parens. Without some sort of distinction being explicitly drawn between points and vectors, more students do fall into the belief that the two are the same thing, or (worse) that each is “the same thing as” a list of numbers in a coordinate representation. Within the context of a course on multivariable calculus, it’s possible to get by with these ideas, but in the long run they will have to be corrected before proceeding into more general contexts.

So, why bring this up now in particular? Because it explains the notation we use in the differential. When we write $df(x;t)$, the semicolon distinguishes between the point variable and the vector variable. It becomes even more apparent when we choose coordinates and write $df(x^1,\dots,x^n;t^1,\dots,t^n)$. Notice that we only ask that $df$ act linearly on the vector variable, since “linear transformations” are defined on vector spaces, not Euclidean spaces.

September 28, 2009 - Posted by | Analysis, Calculus, rants, Topology