3
$\begingroup$

Consider the complex Segre map $s:\mathbb{CP}^1 \times \mathbb{CP}^1 \to \mathbb{CP}^3$ given by $([x_0:x_1],[y_0:y_1]) \mapsto [x_0 y_0: x_0 y_1: x_1 y_0: x_1 y_1]$.

We know that $H^*( \mathbb{CP}^n; \mathbb{Z}) \cong \mathbb{Z}[X]/(X^{n+1})$.

By functorial property $s$ induces map on cohomology groups and therefore a morphism graduated rings (since $H^*(-)$ inherits via $\cup$-product a graded ring structure):

$$s^*:\mathbb{Z}[X]/(X^{4}) \cong H^*( \mathbb{CP}^3; \mathbb{Z}) \to H^*(\mathbb{CP}^1 \times \mathbb{CP}^1; \mathbb{Z}) \cong H^*( \mathbb{CP}^1; \mathbb{Z}) \otimes H^*( \mathbb{CP}^1; \mathbb{Z}) \cong \\ \mathbb{Z}[a]/(a^{2}) \otimes \mathbb{Z}[b]/(b^{2})$$

My question is how to prove that this morphism is concretely given by $X \mapsto a + b$?

Ideas: I know that for coefficients $\mathbb{Z}/2$ instead of $\mathbb{Z}$ to canonical inclusion $i:\mathbb{CP}^{n+1} \to \mathbb{CP}^n$ the corresponding graduiated ring map $i^*:\mathbb{Z}/2[X]/(X^{n+1}) \cong H^*( \mathbb{CP}^n; \mathbb{Z}/2) \to H^*( \mathbb{CP}^{n-1}; \mathbb{Z}/2) \cong \mathbb{Z}[Y]/(Y^{n})$ is given by $X \to Y$.

Therefore (intending to interate this fact) to solve this problem I need to know:

Firstly: If we consider $i^*$ with coefficients in $\mathbb{Z}$, is it also given by $X \mapsto Y$? (If yes, why?)

Secoundly: To use this argument I need "canonical inclusions" $j_i: \mathbb{CP}^1 \hookrightarrow \mathbb{CP}^1 \times \mathbb{CP}^1, i=1,2$, such that $s \circ j_{1/2} = i: \mathbb{CP}^1 \hookrightarrow \mathbb{CP}^3$ coincides with can inclusion. But $\mathbb{CP}^1 \times \mathbb{CP}^1$ is a product so I can't expect that such "inclusions" are given. Can I wlog consider the $\mathbb{CP}^n$ as based spaces?

Can anybody help help/ explain how to prove theese to points?

$\endgroup$
  • $\begingroup$ You can consider $\mathbb{C}P^n$ as a based space. Since it is path connected and simply connected any two choices of basepoints will give you the same answers in your calculations. $\endgroup$ – Tyrone Sep 11 '18 at 10:17
1
$\begingroup$

Firstly lets recall some notation. Let $i:\mathbb{C}P^n\hookrightarrow \mathbb{C}P^{n+k}$ be the canonical inclusion. Then $H^*\mathbb{C}P^n\cong\mathbb{Z}[x_2]/(x_2^{n+1})$ and $H^*\mathbb{C}P^{n+k}\cong\mathbb{Z}[y_2]/(y_2^{n+k+1})$ as graded algebras ($x_2$ and $y_2$ are degree $2$ classes and all cohomology is with integral coefficients). Since $i^*:H^*\mathbb{C}P^{n+k}\rightarrow H^*\mathbb{C}P^n$ is a homomorphism of graded algebras, to determine its action it suffices to determine the value of $i^*y_2$.

Now $\mathbb{C}P^1\cong S^2$ and for $r\geq 1$ the inclusions $\mathbb{C}P^n\hookrightarrow \mathbb{C}P^r$ are $3$-equivalences. You can see this explicitly by studying the standard $CW$ structure on $\mathbb{C}P^r$ and identifying the inclusion with the inclusion of the bottom cell of $\mathbb{C}P^r=S^2\cup_{\eta}e^4\cup\dots\cup e^{2r}$. This tells us that the induced maps $\pi_2\mathbb{C}P^1\xrightarrow{\cong}\pi_2\mathbb{C}P^r$ are isomorphisms, which in turn tells us that the maps $H^2\mathbb{C}P^r\xrightarrow{\cong} H^2\mathbb{C}P^1$ are isomorphisms. Since the digram

$\require{AMScd}$ \begin{CD} \mathbb{C}P^1@>=>> \mathbb{C}P^1\\ @VV V @VV V\\ \mathbb{C}P^n @>i>> \mathbb{C}P^{n+k}. \end{CD}

commutes we may apply the (contravariant) functor $H^2$ and get

$\require{AMScd}$ \begin{CD} H^2\mathbb{C}P^{n+k}@>i^*>> H^2\mathbb{C}P^n\\ @VV =V @VV =V\\ H^2\mathbb{C}P^1 @>=>> H^2\mathbb{C}P^1, \end{CD}

concluding that $i^*:H^2\mathbb{C}P^{n+k}\xrightarrow{\cong} H^2\mathbb{C}P^n$ is an isomorphism. That is, $i^*y_2=x_2$, and more generally

$$i^*(y_2^r)=(i^*y_2)^r=x_2^r.$$

If you don't want to go though all this fuss, then simply observe that $H_*\mathbb{C}P^r$ is torsion free so universal coefficients gives the duality $H^*\mathbb{C}P^r\cong Hom(H_*\mathbb{C}P^r,\mathbb{Z})$ and its not too hard to get integral results by bootstrapping up from mod $p$ calculations.

Now to address your second question. Firstly, I would argue that the inclusions are given, and that the problem should be phrased in the based topological category to begin with. For instance, for $n\geq 0$ it is natural to define the $(n+1)$-sphere as the (reduced) suspension of the $n$-sphere, $S^{n+1}=\Sigma S^n$. Then each sphere gets a canonical basepoint inherited from $S^0$. These basepoints may be taken to correspond exactly to the inclusion of $\mathbb{K}^1$ in $\mathbb{K}^{n+1}$ ($\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$, depending on taste). Then the basepoint of $\mathbb{C}P^n=S^{2n+1}/\mathbb{S}^1=\mathbb{C}^{n+1}/\mathbb{C}^\times$ inherited by the quotient construction is the point

$$[1,0,0,\dots, 0],$$

which is exactly the image of the inclusion of $\ast=\mathbb{C}P^0$.

Since the Segre embedding is derived from the basepoint preserving map $\mathbb{C}^{n+1}\times\mathbb{C}^{n+1}\rightarrow \mathbb{C}^{n+1}\otimes\mathbb{C}^{n+1}\cong\mathbb{C}^{(n+1)^2}$, it is only natural to assume that it too preserves the canonical basepoints in $\mathbb{C}P^n\times\mathbb{C}P^n$ and $\mathbb{C}P^{(n+1)^2-1}$ (you can check that it does).

That aside, there is no difficulty with using basepoints in your problem. Simply choose one, perform your calculation, and then verify that the result is independent of your choice of basepoint.

Now in general, if $X$ is an unbased space, and $x_0\in X$, then the pointed homotopy type $(X,x_0)$ will depend firstly on the path component $X_0$ of $x_0$. For example $\pi_n(X,x_0)=\pi_n(X_0,x_0)$ for $n\geq 1$. Since $\mathbb{C}P^n$ is path connected, the choice of basepoint on this level does not effect anything. If $x_0,x_1\in\mathbb{C}P^n$ are two chosen basepoints then simply choose a path $\gamma_0:I\rightarrow \mathbb{C}P^n$ from $x_0$ to $x_1$ . Since $\mathbb{C}P^n$ is CW, the inclusion of any point $\ast\hookrightarrow \mathbb{C}P^n$ is a closed cofibration, so you can use the homotopy extension property to fiddle with any homotopies and maps you used in calulating with $x_0$ to change your calculation to be with $x_1$ instead.

The next (and last) difficulty is that your resulting maps may depend on the choice of path $\gamma_0$. For example, if $\gamma_1:I\rightarrow \mathbb{C}P^n$ is a second path from $x_0$ to $x_1$, then the concatenation $\gamma_1\ast\gamma_0$ is a closed loop at $x_0$, and thus represents an element in $\pi_1(\mathbb{C}P^n,x_0)$. For general spaces $X$ this loop may be non-trivial, and may affect your conclusions in moving your calculations from $x_0$ to $x_1$ so you must take care.

However, since $\pi_1(\mathbb{C}P^n,x_0)=0$, there is no danger here of that occuring, and you are free to choose and use whatever basepoints in $\mathbb{C}P^n$ and $\mathbb{C}P^{n+k}$ you like without it affecting the outcome of your calculations (although your choices must be consistent among themselves, of course).

If you want to start worrying about homotopies between homotopies, and so on, then the non-trivial higher homotopy groups of $\mathbb{C}P^n$ will start to come into play, but for the problem you have laid out, I do not think we should worry too much about this.

$\endgroup$
  • $\begingroup$ Hi. Thank you for this detailed answer. One point seems unclear: Indeed, considering cell structures one sees that $\mathbb{C}P^n\hookrightarrow \mathbb{C}P^r$ is a $3$-equivalence and therefore $\pi_2\mathbb{C}P^1\xrightarrow{\cong}\pi_2\mathbb{C}P^r$ is an isomorphism. But how do you deduce from this that $H^2\mathbb{C}P^r\xrightarrow{\cong} H^2\mathbb{C}P^1$ is also an iso? Using Hurewicz we get only such result for $H_2\mathbb{C}P^r\xrightarrow{\cong} H_2\mathbb{C}P^1$,right? Do you use implicitely the coefficient theoreme? $\endgroup$ – KarlPeter Sep 11 '18 at 22:41
  • $\begingroup$ @KarlPeter, you can use the Universal Coefficient Theorem for cohomology. Since $H_1\mathbb{C}P^r=0$ it tells you that there is a natural isomorphism $H^2\mathbb{C}P^r\cong Hom_\mathbb{Z}(H_2\mathbb{C}P^r,\mathbb{Z})$. In general $H_*\mathbb{C}P^r$ is torsion free, so this duality holds in all degrees. $\endgroup$ – Tyrone Sep 12 '18 at 9:09
  • $\begingroup$ You can also use cellular cohomology to see this specific result directly. $\endgroup$ – Tyrone Sep 12 '18 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.