On the non-zero elements in $H^*(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)$

Question

The following is an attempt to prove that $x^3$ and $x^4$,respectively, are the non-zero elements in the cohomology groups $H^6(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F} _2)$ and $H^8(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)$. It is imperative to verify the validity of this argument and I welcome any alternative approaches.

The inclusion map $i: \mathbb{C}P^2 \hookrightarrow \mathbb{C}P^4$ induces a restriction map $i^{*}: H^\bullet(\mathbb{C}P^4; \mathbb{F}_2) \to H^\bullet(\mathbb{C}P^2; \mathbb{F}_2).$ Then, in any even degree, we have a short exact sequence $$ 0\to H^{2n}(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)\to H^{2n}(\mathbb{C}P^4; \mathbb{F}_2)\to H^{2n}(\mathbb{C}P^2; \mathbb{F}_2)\to 0.$$ On the other hand, as is known, $H^*(\mathbb{C}P^4; \mathbb{F}_2) \cong \mathbb{F}_2[x]/\langle x^5\rangle,\, x\in H^2(\mathbb{C}P^4; \mathbb{F}_2);$ further $$ H^k(\mathbb{C}P^n; \mathbb F_2) = \left\{\begin{array}{ll} \mathbb F_2 &\mbox{if $0\leq k\leq 2n,$ and $k$ even},\\ 0 &\mbox{otherwise}. \end{array}\right.$$ Hence by the above short exact sequence, one has the isomorphisms: $H^{6}(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)\cong H^{6}(\mathbb{C}P^4; \mathbb{F}_2)$ and $H^{8}(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)\cong H^{8}(\mathbb{C}P^4; \mathbb{F}_2).$ Thus, under these isomorphisms in combination with the fact that $x$ is the generator of $H^2(\mathbb{C}P^4; \mathbb{F}_2),$ it follows that $x^{3}$ and $x^{4}$ are sent to the non-zero elements of $H^6(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)$ and $H^8(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)$, respectively.

Remark: Owing to $x\in H^{2}(\mathbb{C}P^4; \mathbb{F}_2),$ the Cartan formula yields the following expression: $$ \begin{array}{lll} Sq^2(x^3) & =&Sq^0(x^2)\smile Sq^2(x) + Sq^1(x^2)\smile Sq^1(x) + Sq^0(x)\smile Sq^2(x^2)\\ &=& Sq^0(x^2)\smile Sq^2(x) + Sq^1(x^2)\smile Sq^1(x) + Sq^0(x)\smile (Sq^{1}(x))^{2}\\ &=&x^{2}\smile x^{2} = x^{4}\neq 0\ \ \mbox{(since $Sq^1(x)\in H^{3}(\mathbb{C}P^4; \mathbb{F}_2) = 0$)}. \end{array}$$ This leads to a non-trivial homomorphism $$Sq^2: H^6(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)\to H^8(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2).$$

Answer 1

Your argument is flawed. You claim that $x^3$ is in the image of $\delta$, but it's not. The long exact sequence looks like $$ \cdots \to H^5 \mathbb{C}P^4 \to H^5\mathbb{C}P^2 \xrightarrow{\delta} H^6 \mathbb{C}P^4 / \mathbb{C}P^2 \to H^6 \mathbb{C}P^4 \to H^6 \mathbb{C}P^2 \to \cdots $$ Most of the terms here are zero: $$ \cdots \to 0 \to 0 \xrightarrow{\delta} H^6 \mathbb{C}P^4 / \mathbb{C}P^2 \to H^6 \mathbb{C}P^4 \to 0 \to \cdots $$ So no nonzero elements are in the image of $\delta$. Perhaps more usefully, though, there is an isomorphism $$ H^6 \mathbb{C}P^4 / \mathbb{C}P^2 \cong H^6 \mathbb{C}P^4. $$

More generally, all of the odd-dimensional cohomology groups are zero. In any even degree, you get a short exact sequence $$ 0 \xrightarrow{\delta} H^{2m} \mathbb{C}P^4 / \mathbb{C}P^2 \to H^{2m} \mathbb{C}P^4 \to H^{2m} \mathbb{C}P^2 \to 0, $$ either the left term or the right term is zero, depending on the dimension.

Answer 2

Regarding your revised argument:

(a) You need to define $x$ before discussing $x^3$ and $x^4$.

(b) Your first sentence says, "The following is an argument regarding the non-zero elements $x^3$ and $x^4$, respectively, in the cohomology groups $H^6(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F} _2)$ and $H^8(\mathbb{C}P^4/\mathbb{C}P^2; \mathbb{F}_2)$." So you're already asserting that these are the nonzero elements. I thought that this was what you were trying to prove, so you should not assert it. Perhaps instead something like, "The following is an attempt to prove that $x^3$ and $x^4$, respectively, are the non-zero elements in ..."

(c) Furthermore, it is sloppy to say that $x^3$ is the nonzero element in $H^6\mathbb{C}P^4/\mathbb{C}P^2$: $x^3$ is instead an element of $H^6 \mathbb{C}P^4$, and the argument shows that $H^6 \mathbb{C}P^4 \cong H^6\mathbb{C}P^4/\mathbb{C}P^2$. Under this isomorphism, $x^3$ is sent to the nonzero element of $H^6\mathbb{C}P^4/\mathbb{C}P^2$.

($H^*\mathbb{C}P^4/\mathbb{C}P^2$ is a ring, but the nonzero element in $H^6 \mathbb{C}P^4/\mathbb{C}P^2$ is not the cube of any element in that ring, so it should not have the name "$x^3$".)