18

Wedge product, join, etc.,—all of them are no problem for my head, but I am really failing to get a visual idea of what the smash product wants to tell me. For example if I take two spheres, I have no idea what the smash product looks like. The definition is totally clear to me, but I don't know how this thing could eventually look like.

Is anybody here who is good at visualising this or explaining how I could get the idea?

Alex Ortiz
  • 24,844

1 Answers1

20

The smash product of two $1$-spheres gives a nice start in understanding things. You start with $S^1 \times S^1$, a torus. Draw a latitude curve and a longitude curve (i.e., $\{a\} \times S^1$ and $S^1 \times \{b\}$). These intersect at a point $(a, b)$ on the torus, and their union is $S^1 \vee S^1$.

The smash product is the result of collapsing this figure-eight shape to a single point.

It's probably easiest to see this by visualizing the torus as a square with pairwise opposite edges identified. The wedge-product of the two circles then corresponds exactly to the boundary of the square, which each point being seen twice in the square (on opposite sides) except for the join-point, which corresponds to all four vertices.

The smash product then identifies all these boundary points to a single point, i.e., the quotient is a disk with its boundary collapsed to a point, i.e., a 2-sphere.

If I recall correctly, in general $S^k \wedge S^p$ is just $S^{k+p}$; one sort of intuitive argument for the smash-product being a fairly simple space, at least at the homotopy level, is that in the cross product, there's some $\pi_k$ and $\pi_p$, coming from the injection map on each factor, but collapsing the wedge product to a point kills off both of these. More useful, however, is to work the analogy of the 2D example: $S^k \times S^p$ looks like a rectangle in $\mathbb R^{k+p}$ with some identifications; the stuff that ends up identified is exactly the wedge product. If you work this out for $S^1 \times S^2$ in 3-space, I think it'll be fairly clear how it generalizes.

John Hughes
  • 93,729
  • "The smash product then identifies all these boundary points to a single point, i.e., the quotient is a disk with its boundary collapsed to a point, i.e., a 2-sphere."

    Time for a naive question: wouldn't the boundary of a disk collapsed to a single point be... a point? How are you defining the verb "collapse" here?

     – Fomalhaut Mar 23 '16 at 09:43
  • Yes...exactly. So you take a disk and you collapse the boundary to one point. One way to think of this topologically is to take the lower hemisphere of the earth's surface: that's a disk whose boundary is the equator. Now imagine that that lower hemisphere is made of rubber, and gradually stretch it by moving the equator north on the earth's surface, up to the tropic of cancer, then the arctic circle. You've still got a disk, but its boundary circle is now quite small. Keep going, until the circle that was the equator is moved up so far that it all collapses to a single point--the north pole. – John Hughes Mar 23 '16 at 10:43
  • I was thinking for a moment, "but wouldn't the 'filling' disappear with the point?" But then I realized that there's nothing about the boundary that uniquely determines the "filling" of a disk.

    When you have a circle, there seems to be a natural "filling" for the circle, but what if you have some irregular, squiggly closed loop? There are an infinite number of fillings you can assign. So a "disk" must be a closed loop "glued" to some "filling."

     – Fomalhaut Mar 23 '16 at 19:59
  • Good point. "The disk" means $D = { (x, y} \mid x^2 + y^2 \le 1 }$, i.e., a unit disk in the plane. But "a disk" is anything $E$ homeomorphic to this, i.e., for which there's a continuous 1-1 onto map from $D$ to $E$ whose inverse is also continuous. So a filled-in ellipse in the plane, or a filled-in rectangle, or a fattened letter "C" in the plane are all "disks" to a topologist. A "disk" in 3-space is similar, so a partly-inflated soap-bubble on a bubble-wand is also a disk; getting the definition precisely correct took years, so this sort of intuitive description is all I can do for now – John Hughes Mar 23 '16 at 21:16