Timeline for Why is $\mathbb{R}^{\infty}$ defined the way it is?
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| Aug 29, 2011 at 14:43 | comment | added | Stephen | In fact, that's the only sentence in the OP's question with a question mark after it, so I'm even more puzzled by your assertion that you don't see anything that implies we are comparing the direct sum with the direct product. | |
| Aug 29, 2011 at 14:40 | comment | added | Stephen | From the question: "My question is why do we insist that only finitely many of the xi are non-zero for each (x1,x2,x3,…)∈R∞?" If we do not insist this, we get the direct product, do we not? This seems unambiguous to me. | |
| Aug 26, 2011 at 13:32 | comment | added | Andrew Stacey | SPG I didn't (and still don't) see anything that implies that we are comparing the direct sum with the direct product. It seems more "Why the direct sum as opposed to ... err ... anything else?". There certainly are other things that work just as well. (I don't know, either, whether the direct product is one of them.) But that's one of the issues I have with this question: it is possible to read it in too many different ways and each has a subtly different answer. | |
| Aug 24, 2011 at 12:55 | comment | added | Stephen | Andrew, I thought the question was pretty clear (and don't understand all the fuss): why define the Grassmannian of n-planes using the direct sum instead of direct product? The cleanest possible answer is: because using direct product doesn't classify vector bundles on paracompact spaces. Unfortunately, it's not clear to me that this is true, but it might be worth thinking about for a little while for someone who is interested. | |
| Aug 24, 2011 at 11:15 | comment | added | Andrew Stacey | (ctd) I actually think that the last two sentences in your comment are the best answer to this question: they needed a model, and this one is the nearest one to hand. | |
| Aug 24, 2011 at 11:14 | comment | added | Andrew Stacey | SPG: I'm not sure how this question is to be read, which is one of the problems with it as an MO question. I'm not sure what you get for the direct product, but there are many spaces in between which are perfectly fine. As to why they all have the same homotopy type, then that's a theorem on infinite dimensional manifolds which identifies their homotopy types with a colimit construction. But whatever the model, the results that you start with only depend on the homotopy type so don't say anything particular about one model versus any other. (ctd) | |
| Aug 23, 2011 at 13:01 | comment | added | Stephen | Andrew, I'm not sure our understandings of the question are the same. I thought the question was, "Why use the direct sum of countably many real lines instead of the direct product?". Your comment seems to indicate that you think the questioner was asking why Milnor and Stasheff use the particular model they do rather than any other model. I have to admit, I think the answer to your version of the question is easy: because it's technically convenient. After all, if your goal is thms 5.6 and 5.7, getting any model, by hook or by crook, is fine, and the one they use is the obvious one. | |
| Aug 23, 2011 at 12:49 | comment | added | Stephen | Andrew, it's not clear to me that the other definition will have the same homotopy type. Is it clear to you? | |
| Aug 22, 2011 at 18:04 | comment | added | Andrew Stacey | That result only depends on the homotopy type of the Grassmannian, and all the different possibilities have the same homotopy type, so it doesn't address the actual question as to why Milnor and Stasheff use this particular model. | |
| Aug 22, 2011 at 15:33 | history | answered | Stephen | CC BY-SA 3.0 |