Why is $\mathbb{R}^{\infty}$ defined the way it is? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:30:36Z http://mathoverflow.net/feeds/question/73246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is Why is $\mathbb{R}^{\infty}$ defined the way it is? Fly by Night 2011-08-19T22:05:46Z 2011-08-30T01:06:17Z <p>I've been thinking about Grassmannians recently. Think of $\mathbb{R}^k$ as a $k$-dimensional vector space. Let $\text{Gr}_n(\mathbb{R}^k)$ denote the Grassmannian of all $n$-dimensional vector subspaces of $\mathbb{R}^k.$ (This is a compact, Hausdorf topological manifold of dimension $n(k-n)$.) Let</p> <p>$\Gamma^n(\mathbb{R}^k) := \{ (X,v) : X \in \text{Gr}_n(\mathbb{R}^k) \text{ and } v \in X \} .$</p> <p>There's a standard idea of a vector bundle $\pi : \Gamma^n(\mathbb{R}^k) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^k)$ given by $\pi(X,v) := X.$ This bundle has the nice property that lots of other bundles can be realised as sub-bundles of it. There is a more general definition, where we use $\mathbb{R}^{\infty}$ in place of $\mathbb{R}^k$. My question is about why we define $\mathbb{R}^{\infty}$ the way we do. </p> <p>We define $\mathbb{R}^{\infty}$ as the set of infinite sequences $(x_1,x_2,x_3,\ldots)$ where each $x_i \in \mathbb{R}$ and only finitely many of the $x_i$ are non-zero. We identify $\mathbb{R}^k$ with the sequences of the form $(x_1,\ldots,x_k,0,0,\ldots),$ and then topologize $\mathbb{R}^{\infty}$ as the direct limit of the sequence $\mathbb{R}^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3 \subset \ldots$ Then we get the universal bundle $\pi : \Gamma^n(\mathbb{R}^{\infty}) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^{\infty}).$</p> <p>My question is why do we insist that only finitely many of the $x_i$ are non-zero for each $(x_1,x_2,x_3,\ldots) \in \mathbb{R}^{\infty}$? I understand that it gives a countably infinite dimensional vector space, but that's a result of the definition; it doesn't explain why we define it the way we do. I suspect that it's related to the topology, but I don't really know.</p> <p>Edit: The context is the OP is reading Milnor and Stasheff. </p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73262#73262 Answer by Allen Knutson for Why is $\mathbb{R}^{\infty}$ defined the way it is? Allen Knutson 2011-08-20T05:09:09Z 2011-08-20T05:09:09Z <p>This way it's a CW-complex, so studiable by standard tools of algebraic topology (pretty close to Donu's comment).</p> <p>For a striking example of how these things can depend on the definition, look at the infinite unitary groups $U(\infty)$ vs. $U({\mathbb H})$, where the first is defined as $\bigcup U(n)$ and the second as the group of <em>all</em> unitary operators on Hilbert space. Then $U(\infty)$ has very interesting homotopy ($\pi_n(U(\infty)) = 0$ if $2|n$, $= {\mathbb Z}$ if not), whereas $U({\mathbb H})$ is contractible. </p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Answer by SPG for Why is $\mathbb{R}^{\infty}$ defined the way it is? SPG 2011-08-22T15:33:50Z 2011-08-22T15:33:50Z <p>Perhaps the following is so obvious that no one saw fit to mention it in a comment:</p> <p>Let $X$ be a paracompact topological space. With the definition the way it is, homotopy classes of maps from $X$ to the $n$-Grassmannian are in bijection with isoclasses of rank $n$ vector bundles on $X$. (If you haven't got to them yet, this is thms 5.6 and 5.7). This is what really gets used in application to Stiefel-Whitney classes. </p> <p>I don't know what is classified by the other definition of the $n$-Grassmannian you suggest ($n$ dimensional subspaces of the product of infinitely many copies of the reals). But certainly for your purposes (working through Milnor+Stasheff) this is the point. </p> <p>The fact that the usual Grassmannian is a direct limit is used in several technical lemmas in chapter 5, but I have never tried pushing the construction you suggest through. </p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/74008#74008 Answer by Ryan Budney for Why is $\mathbb{R}^{\infty}$ defined the way it is? Ryan Budney 2011-08-29T22:27:44Z 2011-08-29T22:45:08Z <p>There's been quite a bit of discussion of this thread in the meta thread. I'd like to take a stab at answering my interpretation of what the question is getting at. In a sense there's at least two things going on, and that's part of why I think there's been so much discussion. </p> <p>To take a step back from Grassmannians, I'd like to mention why one might be interested in $\mathbb R^\infty$. The Whitney embedding theorem states that every continuous function $f : N \to \mathbb R^k$ can be approximated uniformly by a $C^\infty$-smooth embedding provided $k \geq 2n+1$ where the dimension of $N$ is $n$. Moreover, any two embeddings $N \to \mathbb R^k$ are isotopic provided $k \geq 2n + 2$. So if you wanted to, you could replace the class "$n$-manifolds up to diffeomorphism" with "isotopy classes of $n$-dimensional submanifolds of $\mathbb R^k$ provided $k \geq 2n+2$". </p> <p>A key nice result about the weak topology on $\mathbb R^\infty$ is that any continuous function from a compact space to $\mathbb R^\infty$ has an image in $\mathbb R^k$ for some $k$. So from the perspective of the Whitney embedding theorem above, "$n$-dimensional manifolds up to diffeomorphism" is precisely "$n$-dimensional submanifolds of $\mathbb R^\infty$ up to isotopy". The key thing here is the ambient space is now independent of the dimension of the manifold you're talking about. This is pretty much exactly what's going on with the Grassmannians. </p> <p>Given a vector bundle $p : E \to B$ over a finite-dimensional space $B$ (say a manifold or a CW-complex), there exists a classifying map for the bundle, meaning $p$ is isomorphic to the pull-back of the tautological bundle over $G_{n,k} \equiv G_k(\mathbb R^n)$. $n$ is just some sufficiently large integer. Although people don't state it this way, this theorem is basically <em>the Whitney embedding theorem</em> but for vector bundles. Because it's saying that up to isomorphism, $E$ is a collection of pairs $(b,v)$ where $b \in B$ and $v \in \chi^{-1}(h(b))$, where $\chi : E_{n,k} \to G_{n,k}$ is the tautological bundle over $G_{n,k}$. $h : B \to G_{n,k}$ the classifying map. In a sense we've "embedded" the vector bundle in Euclidean space, well, we've made the <em>fibers</em> as subspaces of Euclidean space. </p> <p>But again, you have the "for some $n$ sufficiently large" thing. And like with manifolds $n$ has an upper bound in terms of the dimension of $B$ (if $B$ is finite dimensional). Since it's sometimes awkward to carry-around these "for sufficiently large $n$" statements, you take the limit space $G_{\infty,k}$ and now your statement is far more clean, because <em>any</em> $k$-dimensional vector bundle over <em>any</em> space $B$ is the pull back of some map $B \to G_{\infty,k}$. The point is that $G_{\infty,k}$ is a <em>universal</em> space -- independent of $B$ or the vector bundle over $B$. The weak topology is exactly what allows us to ensure this happens. </p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/74015#74015 Answer by Yemon Choi for Why is $\mathbb{R}^{\infty}$ defined the way it is? Yemon Choi 2011-08-29T23:51:38Z 2011-08-29T23:51:38Z <p>If we don't insist that all but finitely many entries are non-zero, then we get the (set-theoretic) product of countably many copies of ${\mathbb R}$. If we look at the degree-n Grassmanian of this object, then this is not the object defined in Milnor Stasheff. Hence there exists at least one theorem that is true for the object they define, which is not true for the object we have just defined.</p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/74025#74025 Answer by paul garrett for Why is $\mathbb{R}^{\infty}$ defined the way it is? paul garrett 2011-08-30T01:06:17Z 2011-08-30T01:06:17Z <p>Obviously tastes/opinions vary, but <em>I</em> think <em>some</em> ambiguous, or insufficiently localized, or not publishable-research-y enough, but nevertheless valuable to (a significant demographic of) research mathematicians... In fact, sometimes these questions are exactly the "dumb, non-research" questions that "everyone" (anyway, <em>many</em> people) have asked themselves... and not received a cogent answer.</p> <p>One cliched-but-important (in my opinion) point is the "naive category-theory" explanation of why $\mathbb R^\infty$ is "defined" to be what it is. This does raise the entirely legitimate meta-meta-question of why we "have definitions", and "who is authorized to make them"... to which the easy answer (in my opinion, with some hindsight) is that, not merely must definitions capture the phenomena of important examples (or else the definitions are dumb), but, actually, as it seems to happen very often, re-ordering the "definition...theorem" sequence to "(mapping-)characterization..." rewrites the narrative so that the required/desired property is written in mildly category-theoretic terms, and the technical bit is perhaps proof-of-existence.</p> <p>That is, a (typically, set-theoretic) "definition" is actually just _one_specific_construction_ of an object whose <em>important</em> features (if it exists at all) are completely determined by its interactions with other objects. That is, its characterization is "category theoretic" rather than "set theoretic".</p> <p>(Yes, this is an advertisement for a certain little bit of category theory, though it is not a <em>paid</em> advertisement, insofar as I do not at all advocate <em>formal</em> category theory, nor would I advocate allocating one's personal resources to fretting over reconciliation of set theory and category theory... e.g., Grothendieck's "universes" and large cardinals? Fun, but likely not refering directly to one's original issue...)</p> <p>So, rewriting the question about "why is the definition of $\mathbb R^\infty$ what it is?", we are <em>required</em> to ask what <em>function</em> this thing should have. Well, it is almost immediate that it should be the _ascending_union_ of the $\mathbb R^n$'s. That is, (upon reflection!) it is a (filtered) colimit (a.k.a. "inductive limit"). That is, it should/must/does have certain diagrammatic/mapping properties... as opposed to goofy set-theoretic constructional details.</p> <p>Issues about infinite-dimensional Grassmannians... infinite-dimensional simplicial complexes... do share that basic feature, namely, that there is a mapping property (if only ascending-union sorts of (filtered) colimit properties) that are relevant.</p> <p>Truly, the above viewpoint seems to me to be extraordinarily efficient/effective as explanatory device...</p> <p>(And, one more time, questions that fail to be "documentable research" sometimes are far more interesting and useful to "us" than more focused ones... Of course, this is not a general rule...)</p>