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Products of complexes

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How products are built

The idea of the product may be traced to the image of a stack, which is a simple arrangement of multiple copies of $X$:

Product as a stack.png

More complex outcomes result from attaching to every point of $X$ a copy of $Y$:

Product and columns.png

Example (vector spaces). As an example from linear algebra, what does the following identity mean? $${\bf R} \times {\bf R} = {\bf R}^2 ?$$ We can think of it as if a copy of the $y$-axis is attached to every point on the $x$-axis. Or, we can think in terms of products of sets: $$\hspace{.32in} x\in{\bf R}, y\in{\bf R} \Longrightarrow (x,y)\in{\bf R} \times {\bf R}.\hspace{.32in}\square$$

Generally, for any two sets $X$ and $Y$, their product set is defined as the set of ordered pairs taken from $X$ and $Y$: $$X \times Y := \{(x,y): x\in X, y\in Y\}.$$

Now, it is important to keep in mind that these three sets are just bags of points. How do these points form something tangible? Before we specify the topology, let's consider a few examples of visualization of products.

Example (square). Let $$[0,1] \times [0,1] := {\bf I} \times {\bf I}= {\bf I}^2$$

Square as product.png

You can see how a copy of $Y={\bf I} $ is attached to every point of $X = {\bf I}$, and vice versa. $\square$

The building blocks here are just subsets of ${\bf R}$ and the construction simply follows that of ${\bf R} \times {\bf R}$.

Exercise. Provide a similar sketch for the cube: $${\bf I} \times {\bf I} \times {\bf I} ={\bf I} \times {\bf I}^2 = {\bf I}^3 .$$

Example (cylinder). Consider $${\bf S}^1 \times {\bf I}. $$

Cylinder as product.png

To build the cylinder this way, we place the circle ${\bf S}^1$ on the plane ${\bf R}^2$ and then attach a copy of $[0,1]$, vertically to each of its points. $\square$

In all of these cases, the product fits into ${\bf R}^2$ or ${\bf R}^3$ and is easy to visualize. What if both sets are in ${\bf R}^2$?

Example (torus). A similar procedure for the torus: $${\bf S}^1 \times {\bf S}^1 = {\bf T}^2$$ is impossible. As both copies of ${\bf S}^1$ lie in ${\bf R}^2$, the product would lie in ${\bf R}^4$. One, however, might think of small circles attached, vertically but with a turn, to each point of the large circle on the plane.

Torus as product.png

$\square$

Example (thickening). First, let's observe that we can understand the product $${\bf S}^1 \times {\bf I} $$ not as a cylinder but as a ring (annulus), if we attach the copies of $Y$ in a different manner:

Ring as product.png

The idea is that the product of a space with the segment ${\bf I}$ means “thickening” of the space. As an example, the product $${\bf S}^2 \times {\bf I} $$ is a thickened sphere:

Thick sphere.png

$\square$

We have followed the rule:

  • attach a copy of $Y$ to every point of $X$ (or vice versa). $\\$

Now, the ways these copies are attached aren't identical! We do see how this gluing is changing as we move in $X$ from point to point. The change, however, is continuous. Is continuity enough to pinpoint what the topology of the product is? The example below shows that the answer is No.

Example. Just as with the cylinder, we are attaching segments to the circle but with a gradual rotation. The result is, of course, the Möbius band.

Mobius band as a fiber bundle.png

$\square$

Exercise. Is the above rule violated in this construction?

Products of spaces

Exercise. Prove the following:

  • $\Big( n \text{-point set} \Big) \times \Big(m \text{-point set}\Big) \approx \Big(nm \text{-point set}\Big)$;
  • ${\bf R} \times \{x\} \approx {\bf R}$;
  • ${\bf R}^n \times {\bf R}^m \approx {\bf R}^{n+m}$;
  • ${\bf I} \times {\bf I}\approx {\bf I}^2 =$ square;
  • ${\bf I}^n \times {\bf I}^m \approx {\bf I}^{n+m}$;
  • ${\bf S}^1 \times {\bf I} \approx$ the cylinder;
  • ${\bf S}^1 \times {\bf S}^1 \approx {\bf T}^2 =$ the torus.

Definition. Given two functions, $$f:Z\to X,\ g:Z\to Y,$$ the product function $$f \times g: Z\to X\times Y$$ is given by $$(f\times g)(z):=(f(z),g(z)).$$

Exercise. Even more general is the product of maps $f:A\to X,\ g:B\to Y$. Define it, show that it includes the last definition.

The projections

Just as a subset comes with the inclusion map and the quotient comes with the identification map, the product also comes with a new map -- the projection.

Example. Let's start with the simple projection of the $xy$-plane on the $x$-axis, $$p: {\bf R}^2 \to {\bf R}.$$ It is given by $p(x,y) = x$:

Projection R2R.png

$\square$

Definition. Suppose $X,Y$ are two topological spaces. The projections $$p_X : X \times Y \to X ,\ p_Y : X \times Y \to Y,$$ of $X\times Y$ on $X$ and $Y$ respectively are defined by $$p_X(x,y) := x,\ p_Y(x,y) := y.$$

Example. The projection of the cylinder on the circle -- one can imagine an old chimney collapsing to the ground:

Projection of cylinder on circle.png

The meaning of the projection of torus $p : {\bf T}^2 \to {\bf S}^1$ is not as obvious because it might seem that in order to map it onto the equator you'd have to tear it:

Torus collapse.png

$\square$

Exercise. Sketch the projection of the torus onto one of its meridians.

The projection of the square ${\bf I}^2=[0,1] \times [0,1]$ to the $x$-axis is the restriction of the map $p$ above. In fact, any subset in the plane can be projected to the $x$-axis:

Projections.png

This is the reason why the restrictions of the projection are also called by the same name. The idea is suggested by that of a shadow, such as the one of this arrow:

Projection R3R2.png

Exercise. Describe projections with a commutative diagram. Hint: you'll need to use inclusions.

Products of complexes

If we are able to decompose a topological space into the product of two others, $Z=X\times Y$, we expect $X$ and $Y$ to be simpler than $Z$ and to help us understand $Z$ better. An example is the torus as the product of two copies of the circle.

The construction should be very simple: for two complexes $K,L$, let $$K\times L:=\{a\times b:\ a\in K, b\in L\}.$$ These are pairwise products of every cell of $K$ and every cell in $L$, of all (possible different) dimensions! What's left is to define the product of two cells.

Simplicial complexes have proven to be the easiest to deal with, until now. The problem we encounter is at the very beginning: the product of two simplices isn't a simplex! It's a prism:

Triangle times segment.png

Then, we can't study the product $K\times L$ of simplicial complexes without further triangulation.

Fortunately, cubical complexes have no such problem because the product of two cubes is a cube! In fact, the product of an $n$-cube and an $m$-cube is an $(n+m)$-cube: $${\bf I}^n \times {\bf I}^m = {\bf I}^{n+m}.$$ In other words, if $a$ and $b$ are $n$- and $m$-cells respectively, then $a \times b$ is an $(n+m)$-cell.

Example (segment times segment). Suppose we have two copies of the complex that represents the segment:

Square as product complex.png

Then $K$ has $3$ cells:

  • $0$-cells: $A$ and $B$,
  • $1$-cell: $a$; $\\$

and $L$ has $3$ cells too:

  • $0$-cells: $C$ and $D$,
  • $1$-cell: $b$. $\\$

Now, to find the product of $K$ and $L$, we take a cell $x$ from $K$ and a cell $y$ from $L$, and add their product $x \times y$ to the list. Then $K \times L$ has $3 \times 3=9$ cells:

  • $0$-cells: $A \times B,\ A \times C,\ B \times C,\ B \times D$;
  • $1$-cells: $A \times b,\ B \times b,\ a \times C,\ a \times D$;
  • $2$-cells: $a \times b$. $\\$

What we have is the complex of the square. $\square$

Example (hollow square times segment). More complicated is the product of the complexes of a hollow square and a segment:

Circle times segment.png

We have for $K \times L$:

  • $0$-cells: $8$,
  • $1$-cells: $8$,
  • $2$-cells: $4$. $\\$

This is the finite analog of the topological product of the circle and the segment, ${\bf S}^1 \times {\bf I}$. The result is the cylinder, as expected. $\square$

Exercise. Finish the example.

Back to simplicial complexes. It is in fact easy to build new simplices from old! Consider how adding a new vertex instantly gives as several:

Cone construction dim 1.png

It is simply a matter of adding a new vertex to each list, which defines a certain operation of simplices:

  • $A \cdot C=AC$;
  • $B \cdot C=BC$;
  • $a \cdot C=AB \cdot C=ABC=aC$.

How do we build an $(n+1)$-simplex from an $n$-simplex? Suppose we have $n$ geometrically independent (in general position) points $A_0 , ...,A_n \in {\bf R}^N$ that represent $n$-simplex $$\sigma := A_0 ...A_n.$$ Suppose another vertex $A_{n+1} = Q$ is also available. If the vertex is geometrically independent of the rest, we have a desired $(n+1)$-simplex $$\tau := A_0 ...A_nA_{n+1}.$$ This is called the cone construction.

Cone construction.png

Meanwhile, on the data side, we are simply adding a new element to the list of vertices of the original simplex, as above.

The construction suggests a simple idea of how to define “products” of simplices -- just combine their lists of vertices!

Definition. The join of an $n$-simplex $\sigma = A_0 ... A_n$ and an $m$-simplex $\tau = B_0 ... B_m$ is an $(n+m+1)$-simplex $$\sigma \cdot \tau := A_0 ... A_nB_0 ... B_m.$$

For example, the $3$-simplex is the join of two $2$-simplices:

This operation isn't commutative because the order affects the orientation of the new simplex.

Exercise. Prove: $$\sigma \cdot \tau = (-1)^{(\dim \sigma +1)(\dim \tau +1)}\tau \cdot \sigma.$$

Exercise. Describe the geometric realization of the join by expressing $|a\cdot b|$ in terms of $|a|,|b|$.

Definition. The join of two simplicial complexes is defined as the set of all pairwise joins of the simplices (including the empty simplex) of the two complexes: $$K \cdot L = \{a \cdot b:\ a\in K, b\in L\}.$$

Exercise. Prove that $K \cdot L$ is a simplicial complex.

Example. The circle is the join of two 2-point complexes:

$\square$

Exercise. Represent these simplicial complexes as joins:

  • the segment, the disk, the $n$-ball;
  • the sphere, the $n$-sphere.

Chains in products

What effect does forming the product of two cubical complexes have on the homology? Is the homology group of the product equal to the product of the homology groups? The idea seems to be confirmed by our most popular example of product, the torus: $$H_1({\bf T}^2) = H_1({\bf S}^1 \times {\bf S}^1) \cong H_1({\bf S}^1) \times H_1({\bf S}^1).$$

We need to understand what happens to the chain groups first, in the cubical case.

According to the last subsection, the cells of $K$ are “cross-multiplied” with those of $L$:

  • an $i$-cell in $K$ and a $j$-cell in $L$ are combined to create an $(i+j)$-cell in $K \times L$. $\\$

As linear combinations of cells, the chains of $K$ are also “cross-multiplied” with those of $L$:

  • an $i$-chain in $K$ and a $j$-chain in $L$ are combined to create an $(i+j)$-chain in $K \times L$. $\\$

Consequently, we won't try to compute the $k$th chain groups $C_k(K\times L)$ of the product from the groups $C_k(K)$ and $C_k(L)$ of chains of the same dimension. Instead, we'll look at the complementary dimensions. In other words,

  • $C_k(K\times L)$ is found from the pairwise products of the elements of $C_i(K)$ and the elements of $C_{j}(L)$ for all pairs $(i,j)$ with $i+j=k$. $\\$

Taken together, these correspondences create a function called the cross product: $${\Large \times} :C_i(K) \times C_{j}(L) \to C_{i+j}(K\times L),$$ given by $$(x,y)\mapsto x\times y.$$

Proposition. The cross product is

  • bilinear; i.e., it is linear on either of the two arguments; and
  • natural; i.e., for any cubical maps $f:K\to K',\ g:L\to L'$, we have

$$(f,g)_{\Delta}(a\times b)=f_{\Delta}(a) \times g_{\Delta}(b).$$

Note that this idea will later apply to homology: the $k$th homology $H_k(K\times L)$ is expressed as the sum of pairwise combinations of $H_k(K)$ and $H_{i-k}(L)$ for all $i$ (the Künneth formula).

Example (torus). Let's consider the torus again.

Torus as product with generators.png

The torus itself, a $2$-manifold, is the product of the two circles, $1$-manifolds, and their $1$st homology groups are generated by the $1$-cycles, the fundamental classes, $a$ and $b$. Now, we conjecture that the $2$-cycle $\tau$ of the void of the torus, the fundamental class of this $2$-manifold, is constructed as the product of these two $1$-classes $a,b$.

This reasoning applies to the other homology classes of the torus. The longitude (or the latitude) of the torus, a $1$-manifold, is the product of a circle, $1$-manifold, and a point, $0$-manifold. We conjecture that the $1$-cycle represented by the longitude (latitude) is constructed as the product of the $1$-cycle of the first circle and the $0$-cycle of the second circle. If this is true, the identity we started with was just a coincidence!

Let's work out this example algebraically...

Circle times circle.png

We list the chain groups of these two complexes and their generators: $$\begin{array}{lll} C_0(K)=< A_1,A_2,A_3,A_4 >, &C_1(K)=< a_1,a_2,a_3,a_4 >;\\ C_0(L)=< B_1,B_2,B_3,B_4 >, &C_1(L)=< b_1,b_2,b_3,b_4 >. \end{array}$$ Then, for the product $$M=K\times L,$$ we have: $$\begin{array}{llll} &C_0(M) = < A_i\times B_j & :i,j=1,2,3,4 >;\\ &C_1(M) = < A_i\times b_j, \ a_i\times B_j& :i,j=1,2,3,4 >;\\ \hspace{.2in }&C_2(M) = < a_i\times b_j & :i,j=1,2,3,4 >.&\hspace{.2in }\square \end{array}$$

As we know, the boundary operator of a cubical complex can be defined on the cubes as products of cubes of lower dimension according to this Leibniz-type formula: $$\partial ^M(a \times b) = \partial ^K a \times b + (-1)^{\dim a}a \times \partial ^L b.$$ This formula is extended to the chains and then serves as the boundary operator $$\partial^M: C_k(M) \to C_{k-1}(M)$$ of the product complex $M:=K \times L$. We also know that $$\partial^M\partial^M =0.$$

Example (torus). To confirm this idea, let's compute the boundary operator for the torus $M$: $$\begin{array}{lllll} \partial (A_i\times B_j) &= \partial A_i \times B_j + A_i \times \partial B_j &=0;\\ \partial (A_i\times b_j) &= \partial A_i \times b_j + A_i \times \partial b_j &=A_i \times \partial b_j;\\ \partial (a_i\times B_j) &= \partial a_i \times B_j + a_i \times \partial B_j &=\partial a_i \times B_j;\\ \partial (a_i\times b_j) &= \partial a_i \times b_j + a_i \times \partial b_j . \end{array}$$

What we see is this: $$\begin{array}{lllll} &\ker \partial ^M &= \ker \partial ^L \times \ker \partial ^L ,\\ \hspace{.31in }&\operatorname{Im} \partial ^M &= \operatorname{Im} \partial ^L \times \operatorname{Im} \partial ^L .&\hspace{.31in }\square \end{array}$$

Exercise. Confirm these identities for the torus example.

Exercise. Consider the projections: $$p_K: K \times L \to K,\ p_L: K \times L \to L,$$ and find their chain maps: $$C(K \times L) \to C(K) ,\ C(K \times L) \to C(L).$$

There is an analog of the Leibniz-like formula above for the joins.

Proposition. For two simplicial complexes $\{K,\partial ^K\},\ \{L,\partial ^L\}$, the boundary operator of their join $k\cdot L$ is given by $$\partial ^M(a \cdot b) = \partial ^K a \cdot b + (-1)^{n+1}a \cdot \partial ^L b.$$

Exercise. Prove the proposition.