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Timothy Gowers asks Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to $R^n$? and lists some reasons. The most powerful of these is probably

There are many important examples throughout mathematics of infinite-dimensional vector spaces. If one has understood finite-dimensional spaces in a coordinate-free way, then the relevant part of the theory carries over easily. If one has not, then it doesn't.

I mean sure, but what else? Does anyone know examples of specific vector spaces?

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    $\begingroup$ Because the isomorphism depends on the choice of a basis and there's no canonical choice in general. $\endgroup$ – egreg Dec 26 '16 at 22:34
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    $\begingroup$ Can you please expand further ? :) $\endgroup$ – TripleA Dec 26 '16 at 22:36
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    $\begingroup$ @A.G, that is the sort of "explanation" that is completely useless to someone who is wondering why we consider vector spaces other than $R^n$... The chances of someone not being already convinced of the utility of the general notion of vector spaces to be aware of the covenience of having an abelian category are close to zero. $\endgroup$ – Mariano Suárez-Álvarez Dec 26 '16 at 22:56
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    $\begingroup$ (Moreover, the full subcategory of the category of vector spaces consisting only of those of the form $R^n$, being equivalent to that of finite dimensional vector spaces, is abelian...) $\endgroup$ – Mariano Suárez-Álvarez Dec 26 '16 at 22:58
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    $\begingroup$ Most of this question is copied from here: dpmms.cam.ac.uk/~wtg10/vspaces.html $\endgroup$ – Hans Lundmark Dec 27 '16 at 14:27

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For any integer $k$, the set $M_k$ of complex-differentiable functions $f$ defined on the upper-half plane $\{x+iy: \, y > 0\}$ that satisfy the equations $$f(z+1) = f(z), \; \; f(-1/z) = z^k f(z)$$ and have limit $\lim_{y \rightarrow \infty} f(iy) = 0$ is a vector space over $\mathbb{C}$.

Two specific elements of $M_k$ include the functions $$E_4(z) = 1 + 240 \sum_{n=1}^{\infty} \sigma_3(n) e^{2\pi i n z} \in M_4$$ and $$E_8(z) = 1 + 480 \sum_{n=1}^{\infty} \sigma_7(n) e^{2\pi i nz} \in M_8.$$ Here, $\sigma_k(n)$ is the divisor sum $\sum_{d | n} d^k$.

Assuming that $E_4 \in M_4$ it is rather easy to show that $E_4^2 \in M_8.$

It can be proved that $M_8$ is one-dimensional, so $E_4^2$ is a multiple of $E_8$. Comparing constant coefficients tells you that they must be equal, and comparing the others gives you the formula $\sigma_7(n) = \sigma_3(n) + 120 \sum_{m=1}^{n-1} \sigma_3(m) \sigma_3(n-m).$

For example $$\sigma_7(2) = 1 + 2^7 = 1 + 2^3 + 120$$ and $$\sigma_7(3) = 1 + 3^7 = 1 + 3^3 + 120(1+2^3 + 1 + 2^3).$$

A lot of vector spaces like this show up in number theory. They are typically finite-dimensional but working out a basis is pretty hard (certainly harder than showing that they are finite-dimensional).

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    $\begingroup$ Very good example. It might be an idea to also mention that these are called modular forms in case someone wants to learn more. $\endgroup$ – Tobias Kildetoft Dec 26 '16 at 23:03
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    $\begingroup$ That formula is magic. $\endgroup$ – Akiva Weinberger Dec 26 '16 at 23:46
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    $\begingroup$ From a perspective of a person who mostly do analysis, I never knew that this beautiful thing exists. Definitely a +1. $\endgroup$ – BigbearZzz Dec 27 '16 at 14:44
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If you decided that you are only going to call "vector space" those of the form $\mathbb R^n$, then you find yourself in the position that now subspaces are no longer vector spaces.

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  • $\begingroup$ Well every strict subspace is still isomorphic to $\mathbb R ^k$ for some $k<n$ $\endgroup$ – ASKASK Dec 27 '16 at 0:00
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    $\begingroup$ But they are not of that form. To be able to even make that statement, @Askask, you need to have an abstract notion of vector spaces. $\endgroup$ – Mariano Suárez-Álvarez Dec 27 '16 at 0:03
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    $\begingroup$ @Mehrdad - that is because you missed the very good insight it gives. In $\Bbb R^3$, every plane through the origin forms a 2-dimensional subspace. There are infinitely many such planes. But only two of them can be directly identified with $\Bbb R^2$. All the others are among the many. many examples of vector spaces without a canonical basis. $\endgroup$ – Paul Sinclair Dec 27 '16 at 3:18
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    $\begingroup$ @PaulSinclair: The question said "if they are all isomorphic to $\mathbb{R}^n$", not "if they are all $\mathbb{R}^n$"... $\endgroup$ – user541686 Dec 27 '16 at 3:20
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    $\begingroup$ @Mehrdad - That was just the lead-in information framing the question, not the question itself. The question was why it isn't enough to just study $\Bbb R^n$ instead of abstract vector spaces. The reason given in this answer is that even when you just study $\Bbb R^n$, you still find vector spaces that are not naturally isomorphic to $\Bbb R^k$ for some $k$. There isn't just one isomophism, or even a special one. To apply your $\Bbb R^n$ knowledge requires arbitrary choices on your part. Those choices actually make things more complicated, not less. $\endgroup$ – Paul Sinclair Dec 27 '16 at 3:35
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Consider an analogous question:

Why consider finite sets in the abstract if they're all isomorphic to $\{1,\ldots,n\}$ for some $n$?

  • Because there could be names for the elements that are more natural for a given situation than $1,\ldots,n$, e.g. we may want to refer to $$\{\text{red},\text{green},\text{blue}\}$$ instead of $$\{1,2,3\}\text{ where we agree that 1 stands for red, 2 for green, 3 for blue}$$

  • In general, names for elements are not always important

  • There are many subsets of a set of the form $\{1,\ldots,n\}$ for some $n$ that are not themselves sets of the form $\{1,\ldots,n\}$ for some $n$

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    $\begingroup$ Your second bullet point seems to contradict your first. I think I get what you mean to say, but it could probably be worded better. $\endgroup$ – Will R Dec 27 '16 at 2:14
  • $\begingroup$ @WillR: the point and its wording seem fine to me. They’re describing two different situations: sometimes, there are natural names that you don’t want to change; other times, names aren’t helpful at all. $\endgroup$ – Peter LeFanu Lumsdaine Dec 28 '16 at 10:57
  • $\begingroup$ @PeterLeFanuLumsdaine: If the fact that the points reference different situations are important, then that fact can be made clear to the reader, and doing so improves the wording of the point. $\endgroup$ – Will R Dec 28 '16 at 16:31
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The cheeky answer is that we would not know that all finite-dimensional vector spaces are isomorphic to $\mathbb{R}^n$, if we did not study finite-dimensional vector spaces in their own right. In mathematics, we generally like to use as few assumptions as possible and to isolate them in the form of axioms.

See https://en.wikipedia.org/wiki/Examples_of_vector_spaces for examples of vector spaces that seem very different from those found in the world of geometry. Function spaces are good examples; the space $X \rightarrow \mathbb{R}$ of all continuous functions from a given topological space $X$ to $\mathbb{R}$ is a natural example.

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  • $\begingroup$ Also, by using as few assumptions as possible, that gives a better starting point for asking whether we can use even fewer and hence if a result can be generalised to a broader class. $\endgroup$ – Keith Dec 27 '16 at 7:12
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I would claim we study them precisely because they are isomorphic to $\mathbb{R}^n$. What do I mean?
I mean that since we are already familiar with $\mathbb{R}^n$, we can use this intuition to understand vector spaces in general, and once we do that, we can generalize the concepts to other less-intuitive objects (such as infinite-dimensional vector spaces) while carrying over our understanding.

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    $\begingroup$ Well said. Finite-dimensional real vector spaces arise in other contexts, for example the set of solutions of a homogeneous differential (or difference) equation. So it helps to have some tools ready to deal with them: the vector-space properties of $\mathbb R^n$. $\endgroup$ – DanielWainfleet Dec 29 '16 at 2:02
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Consider $V$, the space of polynomials of degree less than $3$, which is isomorphic to $\mathbb{R}^3$ via $$ a_0+a_1x+a_2x^2\mapsto \begin{bmatrix} a_0-a_2\\ a_0-a_1-2a_2\\ a_2 \end{bmatrix} $$ Would you recognize what is the linear map $f\colon V\to V$ whose matrix with respect to this isomorphism is $$ \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & -2 \\ 0 & 0 & 1 \end{bmatrix} $$ without looking at the spoiler below?

The linear map can be easily expressed as $p(x)\mapsto p(x)+p'(x)$.

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There are lots of ways that viewing a finite dimensional vector space in its "intrinsic form" is useful, especially in calculus. Here's a simple example. Consider the space of $n \times n$ matrices with real coefficients $\mathbb{M}^n$. We could just regard this as the space $\mathbb{R}^{n^2}$ and do calculus as per usual in higher dimensions.

This does not work so well when we want to consider derivatives of simple maps. For instance, let's look at the map $f: \mathbb{M}^n \to \mathbb{M}^n$ given by $f(M) = M^2$. This is a smooth map, and we can compute that the first derivative satisfies $$ \langle Df(M),N \rangle = MN + NM $$ for all $N \in \mathbb{M}^n$. In this context we view $Df(M) \in \mathcal{L}(\mathbb{M}^n)$, i.e. as a linear map from $\mathbb{M}^n$ to itself. If we insist on throwing away this abstract formulation then we only have $\mathbb{R}^{n^2}$ to work with, in which case there is no product structure. We will still find that this map $f$, viewed as a map from $\mathbb{R}^{n^2}$ to itself, is smooth, but the formula for the derivative will be significantly worse, and even challenging to write down.

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All Hilbert spaces are unitarily isomorphic to some $\ell^2(E)$, but that does not mean we don't study Hilbert spaces.

The important thing is finite-dimensional vector spaces may carry additional structure (such as the space of polynomials) which we are interested in.

Another thing is people occasionally encounter non-trivial subspaces of $\mathbb R^n$, such as a line or a plane. In this case the theory of finite-dimensional vector spaces becomes useful.

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    $\begingroup$ Sort of related, sometimes the isomorphisms between the spaces contain information (like the existence of Brownian Motion by the Weiner isometry.) $\endgroup$ – 3-in-441 Dec 26 '16 at 22:38
  • $\begingroup$ Do you have a link or maybe a specific example where the theory applies? $\endgroup$ – TripleA Dec 26 '16 at 22:40
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Even studying $\mathbb{R}^n$ in the abstract is useful; sweeping irrelevant details under the rug makes the theory a lot cleaner. You can see what's going on more easily if you're not bogged down in coordinates!

This can also be true for actual computation too. Vector calculus and matrix algebra are important tools for doing calculations; one must be able to form a concept of such things as mathematical objects in their own right, rather than merely being arrays of scalar quantities.

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A very concrete example of why we should not stop studying something once we have isomorphisms, although going away from vector spaces.

A lot of modern (asymmetric) cryptography is based on discrete logarithm problems in prime order groups. All these groups are isomorphic to $\mathbb{Z}/q\mathbb{Z}$ for some prime $q$. The two most common examples are probably

  1. A subgroup $G$ of the multiplicative subgroup $\mathbb{F}_p^*$ of a finite field $\mathbb{F}_p$, say of prime order $q$;
  2. The group of rational points $E(\mathbb{F}_p)$ on an elliptic curve $E$, also of prime order $q$.

Note that $G\cong \mathbb{Z}/q\mathbb{Z}\cong E(\mathbb{F}_p)$. As abstract groups, one could think of the discrete logarithm problem to be equivalent in all of them. If we forgot about any structure except for being a group, this is true, as one can show that we need at least $O(\sqrt{q})$ group operations.

However, in practice there is additional structure. If we consider the set of integers modulo $q$, the discrete logarithm is quite trivial. In the group $G$ it is a bit harder, and if we set $\log_2{q}\approx 1000$ then it is currently infeasible (at least has never been publicly done). If we set $\log_2{q}\approx 256$, then the discrete logarithm problem in $G$ becomes feasible, while it is still very hard in $E(\mathbb{F}_p)$.

Groups can carry a lot of additional structure which we do not want to forget about, and I imagine the same to be true for vector spaces. For example, a finite field is a vector space over some prime order field, but is much more than that.

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Some examples of finite-dimensional vector spaces that are useful in my field of practice, convex optimization:

  • real or complex scalars (yes, $\mathbb{R}^1/\mathbb{C}^1$)
  • $\mathbb{C}^n$, the space of complex vectors
  • the space of $m\times n$ real or complex matrices
  • $\mathcal{S}^n$, the space of $n\times n$ symmetric matrices
  • $\mathcal{H}^n$, the space of $n\times n$ Hermitian matrices
  • other spaces of matrices with specific structure; e.g., Toeplitz matrices, Hankel matrices
  • Cartesian products of two or more of the above. For example, a tuple consisting of a symmetric matrix, a vector, and a scalar.

Knowing that these vector spaces are isomorphic to $\mathbb{R}^n$ simplifies the study of convex analysis and convex optimization tremendously, because most of results indeed deriving only once---with appeals to the isomorphism used to quickly apply those results to arbitrary finite-dimensional spaces.

How inconvenient would it be for us to avoid discussion of these other vector spaces? Well, let's consider a simple example. An important function in semidefinite programming is the maximum eigenvalue function on symmetric matrices: $$f:\mathcal{S}^n\rightarrow\mathbb{R}, \qquad f(X) = \lambda_{\max}(X) = \max_{v\neq 0} (v^TXv)/(v^Tv)$$ This a convex function of its input $X$, and as a result we can build practically useful and tractable optimization problems that incorporate it into objective functions or constraints. (Other matrix functions, such as the minimum eigenvalue, the logarithm of the determinant, etc., arise frequently as well.)

Now, because of the isomorphism, we know that there is another function $g:\mathbb{R}^{n(n+1)/2}\rightarrow\mathbb{R}$ that could be used to represent $f$. In other words, to compute $g(x)$, we'd take the $n(n+1)/2$ values of $x$, arrange them in the lower triangle of a matrix, copy the off-diagonal elements to the upper triangle to symmetrize things, then compute the eigenvalue of that matrix.

Whew! Do I really have to do that every time I want to work with a function defined on a non-$\mathbb{R}^n$ vector space? I hope not. Sure, sometimes it might be easier to prove a particular technical result if I do; but no, most of the time I want to work in the natural vector space the problem suggests. Knowing how the principles of vector spaces apply to spaces beyond $\mathbb{R}^n$ enables me to avoid constant translation to and fro.

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