Degrees of maps in algebraic topology

Question

Please can I have some tips on how to construct maps between topological spaces of a given degree? For example, how would you go about building a map of degree $3$ from $\mathbb{CP}^1\times\mathbb{CP}^2 \to \mathbb{CP}^3$? Or a map from $S^2\times S^2 \to \mathbb{CP}^2$ of even degree? I don't know where to start. Are there any particular techniques that are useful?

One idea that comes to mind is to start with a map of nonzero degree, and pre compose it or post compose it with a map of nonzero degree. There are easily described degree $n$ maps $S^2\to S^2$, so if you can find a map $S^2\times \Bbb CP^2\to\Bbb CP^3$ of nonzero degree $d$ (resp. $S^2\times S^2\to\Bbb CP^2$) you should get maps of degree $nd$ for all $n\in\Bbb Z$. EDIT. I think I may have misread your question. — Olivier Bégassat, May 25 '14 at 08:25

score 1 · Answer 1 · answered Jun 11 '17 at 16:19

In general it is hard to write down maps of a given degree, or even to determine whether such maps exist.

I don't know if there is a map $\mathbb{CP}^1\times\mathbb{CP}^2 \to \mathbb{CP}^3$ of degree three, but there are maps $S^2\times S^2 \to \mathbb{CP}^2$ of even degree. In fact, every map $S^2\times S^2 \to \mathbb{CP}^2$ has even degree.

To see this, recall that $H^*(S^2\times S^2; \mathbb{Z}) \cong \mathbb{Z}[\alpha, \beta]/(\alpha^2, \beta^2)$ and $H^*(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}[\omega]/(\omega^3)$. Consider a map $f : S^2\times S^2 \to \mathbb{CP}^2$. Note that $f^*(\omega^2) = (f^*\omega)^2$ and since $f^*\omega \in H^2(S^2\times S^2; \mathbb{Z}) \cong \mathbb{Z}\alpha\oplus\mathbb{Z}\beta$, $f^*\omega = x\alpha + y\beta$ for some $x, y \in \mathbb{Z}$. Therefore

$$f^*(\omega^2) = (f^*\omega)^2 = (x\alpha + y\beta)^2 = 2xy\alpha\beta.$$

As $\alpha\beta$ is a generator of $H^4(S^2\times S^2; \mathbb{Z})$, $\deg f = \pm 2xy$ which is even (the sign depends on the orientations).

score 1 · Answer 2 · answered May 25 '14 at 08:48

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Welcome user48617,

Are you familiar with the Künneth-Theorem and cellular homology?

In order to construct maps that start in $\mathbb CP^1 \times \mathbb CP^2$ or $S^2 \times S^2$ it suffices to construct maps that start in one of the factors using the cell complexes of the respective spaces.

By Künneth Theorem, you get an induced homomorphism of short exact sequences but here the $\mathrm{TOR}$-term vanishes, i.e. the degree of the map in question is in fact the product of the degrees of the simpler maps.

answered May 25 '14 at 08:48

Felix Boes

331

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This looks like a very useful technique-but could you clarify you you go from a map on each coordinate to a map on the product space? So in $S^2\times S^2$ I have two 2-cells and a 4-cell. Given two maps $S^2\to \mathbb{C}P^2$ I can certainly extend to the 2-skeleton of the product, but where do I send the 4-cell? I suppose some obstruction theory would help, but is there an elementary approach? – Kevin Carlson May 25 '14 at 09:41
Thanks Felix! In the particular example of CP1×CP2 to CP3, I don't think your method as stated would work? For dimension reasons there are no maps of non zero degree from CP1 to CP3, for example. Have I misunderstood your answer? – Jeremy Judge May 26 '14 at 07:32

Kevin Carlson · Answer 3 · 2014-05-25T09:53:11.243

For maps involving projective spaces, an elementary technique is to try to find a surjection involving homogeneous polynomials-you have coordinates in this case, so you should use them. Given polynomial maps $p:\mathbb{C} P^1\to \mathbb{C} P^3$ and $q:\mathbb{C} P^2\to \mathbb{C} P^3$ I can construct $pq$ by multiplying coordinatewise: adding wouldn't preserve well-definedness, but multiplying does, as in $([z:w],[a:b:c])\mapsto [za^2:wb^2:zc^2:w(a^2+2b^2+3c^2)].$ A similar technique might do the $S^2\times S^2\to \mathbb{C} P^2$ case, since that's also a map out of a product of projective spaces.

Degrees of maps in algebraic topology

3 Answers3