# Computing product spaces?

1. Feb 14, 2008

### homomorphism

I'm trying to get an intuitive feel for product spaces, and I think I am having some trouble. I understand that the product space S1 x S1 is a torus. the two circles are cycles which are not boundaries though. What happens when you take the product space of two cycles which are boundaries? For example, the product space of two 2-simplices. Is there any "intuitive" way to understand what will happen when you multiply two simplices?

2. Feb 15, 2008

### mrandersdk

maybe you can think of it like, if you take R x S^1, then you place at each point on the real line a copy of S^2, if you draw this it is a cylinder. But but if you fx. take R^2 and place a copy of the cirkel at each point, it's gonna be a mess (try to draw it , so there is not always a nice way to picture product spaces.

3. Feb 15, 2008

### homomorphism

hmm...how about if I have two 1-simplices: [a,b] and [c,d]

If I wanted to compute the product space, would it be two 2-simplices, [a,b,c] and [a,b,d]
or would it be a bunch of 1 simplices: [a,c], [a,d], [b,c], [b,d] ??

4. Feb 15, 2008

### Hurkyl

Staff Emeritus
Actually, $\mathbb{R}^2 \times S^1$ isn't hard: it's the interior of a torus. Try taking a copy of the circle, and attaching a copy of R^2 at each point. (More specifically, use a disk, which is homeomorphic to R^2)

5. Feb 17, 2008

### mrandersdk

okay that was a smart trick, so R^2 x S^1 is homeomorphic to the inside of a torus, didn't know that, but you have to give that the space S^1 x R^2 , even though it is homeomorphic to the inside of a torus, is hard to imagine (in a usefull way), if you don't are comfortable with spaces being homeomorphic.

To answer homomorphism: If simplices just are an closed line from a to b, in R, then you get:

You place a line of length [c,d] orthogonal on the line [a,b] at each point on [a,b], then what do you get?

you get the square with coners (a,c),(a,d),(b,c) and (b,d), remember that

[a,b] subset of R and [c,d] subset of R so [a,b] x [c,d] subset of R x R = R^2

6. Feb 17, 2008

### WWGD

"okay that was a smart trick, so R^2 x S^1 is homeomorphic to the inside of a torus, didn't know that, but you have to give that the space S^1 x R^2 , even though it is homeomorphic to the inside of a torus, is hard to imagine (in a usefull way), if you don't are comfortable with spaces being homeomorphic."

I used to be uncomfortable with this too, until I realized that every metric space
has a (top-) equivalent bounded metric. It is this that allows the (countable) product of
metric spaces to be metrizable. Then you can imagine IR^2 in its "model" as a
ball, like Hurkyl said. Maybe that will help you see IR^2, or IR^n as a finite ball.

7. Feb 17, 2008

### WWGD

Actually, you have to be careful on how you define the product, otherwise you get
a "hole-less" doughnut. I think the product has to be a topological product (equiv.
to a topological or disjoint union) , otherwise, S^1xS^1 will not have a hole:

take two copies S^1 and S^1' , and do the product S^1xS^1' : take x, y, x=/y in

S^1 . Then (x,S^1') and (y,S^1') overlap , so that there is no center hole.

IOW: (x,S^1') will contain a line segment of length 1 ; the point (x,1) in (x,S^1')

and (y,S^1') will also contain a point (y,1).

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