Tags: #spectral-sequences

Computation of \(H^*({\mathbf{CP}}^2)\)

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Theorem

Suppose $F

\to 

\to 

Then there exists a spectral sequence \(E_*\) such that

• \(E_2^{p,q} = H^p(B, H^q(F;{\mathbf{Z}})) = H^p(B;{\mathbf{Z}}) \otimes H^q(F;{\mathbf{Z}})\)
• \(E_\infty^{p,q} \Rightarrow H^{p+q}(E)\)

Computation

Use the above theorem with the fibration \(S^1 \to S^5 \to{\mathbf{CP}}^2\), as well as the following facts:

• \(H^*(S^1) = {\mathbf{Z}}\delta_0 + {\mathbf{Z}}\delta_1\)
• \(H^*(S^5) = {\mathbf{Z}}\delta_0 + {\mathbf{Z}}\delta_5\)
• \(H^0({\mathbf{CP}}^2) = {\mathbf{Z}}\) (i.e. it is simply connected)
• \(d_2: E_2^{p,q} \to E_2^{p-2, q+1}\)

By the theorem, we have

\begin{align*}E_2^{p,q} = H^p({\mathbf{CP}}^2) \otimes H^q(S^1)\end{align*}

Thus the \(E_2\) page of the spectral sequence looks like this:

latex {cmd=true hide=true run_on_save=true} \documentclass{standalone} \usepackage{tikz} \usepackage{dsfont} \usepackage{amsmath, amsthm, amssymb} \usetikzlibrary{matrix} \newcommand*\Z{\mathds{Z}} \newcommand*\CP{\mathbb{CP}} \newcommand*\ee[2]{H^{#1}(\CP^2) \otimes_{\Z} H^{#2}(S^1)} \newcommand*\zt[2]{#1 \otimes_{\Z} #2} \newcommand*\HT[2]{H^{#1}(\CP^2) \otimes_{\Z} #2} \newcommand*\HCP[1]{H^{#1}(\CP^2)} \begin{document} \begin{tikzpicture} \matrix (m) [matrix of math nodes, nodes in empty cells,nodes={minimum width=5ex, minimum height=5ex,outer sep=-5pt}, column sep=1ex,row sep=1ex]{ % S^1& & & & & & & & \\ % 0& \ee{0}{5}& \ee{1}{5}& \ee{2}{5}& \ee{3}{5}& \ee{4}{5}& \ee{5}{5}& &\\ 0& \ee{0}{4}& \ee{1}{4}& \ee{2}{4}& \ee{3}{4}& \ee{4}{4}& \ee{5}{4}& &\\ 0& \ee{0}{3}& \ee{1}{3}& \ee{2}{3}& \ee{3}{3}& \ee{4}{3}& \ee{5}{3}& &\\ 0& \ee{0}{2}& \ee{1}{2}& \ee{2}{2}& \ee{3}{2}& \ee{4}{2}& \ee{5}{2}& &\\ 0& \ee{0}{1}& \ee{1}{1}& \ee{2}{1}& \ee{3}{1}& \ee{4}{1}& \ee{5}{1}& &\\ 0& \ee{0}{0}& \ee{1}{0}& \ee{2}{0}& \ee{3}{0}& \ee{4}{0}& \ee{5}{0}& &\\ \quad\strut % & 0& 1& 2& 3& 4& 5& \CP_2& \strut \\}; % \draw[thick] (m-8-1.east) -- (m-1-1.east) ; \draw[thick] (m-8-1.north) -- (m-8-9.north) ; \end{tikzpicture} \end{document}

Using the facts above, we can substitute in some known groups:

latex {cmd=true, hide=true, run_on_save=true} \documentclass{standalone} \usepackage{tikz} \usepackage{dsfont} \usepackage{amsmath, amsthm, amssymb} \usetikzlibrary{matrix} \newcommand*\Z{\mathds{Z}} \newcommand*\CP{\mathbb{CP}} \newcommand*\ee[2]{H^{#1}(\CP^2) \otimes_{\Z} H^{#2}(S^1)} \newcommand*\zt[2]{#1 \otimes_{\Z} #2} \newcommand*\HT[2]{H^{#1}(\CP^2) \otimes_{\Z} #2} \newcommand*\HCP[1]{H^{#1}(\CP^2)} \begin{document} \begin{tikzpicture} \matrix (m) [matrix of math nodes, nodes in empty cells,nodes={minimum width=5ex, minimum height=5ex,outer sep=-5pt}, column sep=1ex,row sep=1ex]{ % S^1& & & & & & & & \\ % 0& \zt{\Z}{0}& \HT{1}{0}& \HT{2}{0}& \HT{3}{0}& \HT{4}{0}& \HT{5}{0}& &\\ 0& \zt{\Z}{0}& \HT{1}{0}& \HT{2}{0}& \HT{3}{0}& \HT{4}{0}& \HT{5}{0}& &\\ 0& \zt{\Z}{0}& \HT{1}{0}& \HT{2}{0}& \HT{3}{0}& \HT{4}{0}& \HT{5}{0}& &\\ 0& \zt{\Z}{0}& \HT{1}{0}& \HT{2}{0}& \HT{3}{0}& \HT{4}{0}& \HT{5}{0}& &\\ 0& \zt{\Z}{\Z}& \HT{1}{\Z}& \HT{2}{\Z}& \HT{3}{\Z}& \HT{4}{\Z}& \HT{5}{\Z}& &\\ 0& \zt{\Z}{\Z}& \HT{1}{\Z}& \HT{2}{\Z}& \HT{3}{\Z}& \HT{4}{\Z}& \HT{5}{\Z}& &\\ \quad\strut % & 0& 1& 2& 3& 4& 5& \CP_2& \strut \\}; % \draw[thick] (m-8-1.east) -- (m-1-1.east) ; \draw[thick] (m-8-1.north) -- (m-8-9.north) ; \end{tikzpicture} \end{document}

Now recalling two useful properties of the tensor product:

• \(A \otimes_{\mathbf{Z}}0 = 0\), and
• \(A \otimes_{\mathbf{Z}}{\mathbf{Z}}= A\),

we obtain the following simplified version of the \(E_2\) page, with several of the potentially non-trivial differentials indicated:

```latex

\begin{document}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes,
  nodes in empty cells,nodes={minimum width=5ex,
  minimum height=5ex,outer sep=-5pt},
  column sep=1ex,row sep=1ex]{
%
S^1&    &   &   &   &   &   &   &   \\
%
5&  0&      0&  0&  0&  0&  0&  &\\
4&  0&      0&  0&  0&  0&  0&  &\\
3&  0&      0&  0&  0&  0&  0&  &\\
2&  0&      0&  0&  0&  0&  0&  &\\
1&  \Z& \HCP{1}&    \HCP{2}&    \HCP{3}&    \HCP{4}&    \HCP{5}&    \cdots&\\
0&  \Z& \HCP{1}&    \HCP{2}&    \HCP{3}&    \HCP{4}&    \HCP{5}&    \cdots&\\ \quad\strut
%
&   0&  1&  2&  3&  4&  5&  {\mathbf{CP}}_2& \strut \\};
%
\draw[thick] (m-8-1.east) -- (m-1-1.east) ;
\draw[thick] (m-8-1.north) -- (m-8-9.north) ;
\draw[-stealth] (m-6-2.south east) -- (m-7-4.north west);
\draw[-stealth] (m-6-3.south east) -- (m-7-5.north west);
\draw[-stealth] (m-6-4.south east) -- (m-7-6.north west);
\draw[-stealth] (m-6-5.south east) -- (m-7-7.north west);
\draw[-stealth] (m-6-6.south east) -- (m-7-8.north west);

\end{tikzpicture}
\end{document}

Now we use the fact that the spectral sequence converges to make several deductions:

Claim:

$$H^1(S^5) = 0 \implies H^2({\mathbf{CP}}^2) \cong {\mathbf{Z}}~\text{and}~H^1({\mathbf{CP}}^2) = 0$$

(This will be a template argument for most of the rest, so I will spell out more details here and gloss over them later.)

• This means that \(E_\infty^{0,1} \oplus E_\infty^{1,0} = 0\).
• Consider the process of obtaining the \(E_3\) page:
  • \(E_3^{0,1}\) is obtained from the homology of the complex \(0 \to{\mathbf{Z}}\xrightarrow{{\partial}_1} H^2({\mathbf{CP}}^2) \to 0\), i.e. we have \(E_3^{0,1} = \frac{\ker {\partial}_1}{\operatorname{im}0} = \ker {\partial}_1\)
    • Note that all differentials after the \(E_3\) page extend into the \(p<0\) and \(q<0\) quadrants, so there is stabiization here and \(E_3^{0,1} = E_\infty^{0,1}\)
      • But if the homology of this sequence is not zero, then \(E_3^{1,0} \neq 0\), so \(E_\infty^{0,1} \neq 0\) and \(E_\infty^{0,1} \oplus E_\infty^{1,0} \neq 0\), a contradiction.
    • So this is an acyclic complex, and thus an exact sequence.
    • So \({\partial}_1\) is an isomorphism, and \(H^2({\mathbf{CP}}^2) \cong {\mathbf{Z}}\)
  • \(E_3^{1,0}\) is obtained from the homology of \(0 \to H^1({\mathbf{CP}}^2) \to 0\)
    • By the same argument, this spot stabilizes at \(E_3\) and so this complex must have trivial homology.
    • But this can only happen if \(H^1({\mathbf{CP}}^2) = 0\)

Claim:

$$H^2(S^5) = 0 \implies H^1({\mathbf{CP}}^2) \cong H^3({\mathbf{CP}}^2) ~\text{and}~ H^2({\mathbf{CP}}^2) = {\mathbf{Z}}$$

We have \(H^2(S^5) = E_\infty^{0,2} \oplus E_\infty^{1,1} \oplus E_\infty^{2,0}\).

Note that \(E_2^{0,2} = 0\), so \(E_\infty^{0,2} = 0\) there are only two contributing terms to consider.

\(E_\infty^{1,1}\): This involves looking at the complex \(0 \to H^1({\mathbf{CP}}^2) \xrightarrow{{\partial}_2} H^3({\mathbf{CP}}^2) \to 0\). All differentials extend into zero quadrants starting at \(E_3\), so this entry stabilizes at \(E_3\). But any homology in this complex would contribute a nonzero contribution to \(H^2(S^5)\), so this complex is acyclic/exact and \({\partial}_2\) is an isomorphism.

\(E_\infty^{2,0}\): This involves \(0 \to{\mathbf{Z}}\xrightarrow{f} H^2({\mathbf{CP}}^2) \to 0\), where the \(E^3\) differentials extend into zero quadrants and thus this entry stabilizes at \(E^3\). Any nonzero homology here yields a nonzero contribution to \(H^2(S^5)\), so this complex is acyclic/exact and thus \(f\) is an isomorphism.

Claim:

$$H^3(S^5) = 0 \implies H^2({\mathbf{CP}}^4) \cong H^4({\mathbf{CP}}^2) \cong H^6({\mathbf{CP}}^2),~ H^1({\mathbf{CP}}^2) \cong H^3({\mathbf{CP}}^2) \cong H^5({\mathbf{CP}}^2)$$

Note: this is the first spot where the differentials may not extend into zero quadrants, but since the total homology is zero, this is not a real issue yet.

We have \(H^3(S^5) = \displaystyle\bigoplus_{p+q = n}E_\infty^{p,q} = E_\infty^{0,3} \oplus E_\infty^{1,2} \oplus E_\infty^{2,1} \oplus E_\infty^{3,0}\). Every summand must be zero, so we examine them individually.

\(E_\infty^{0,3}\): We have \(E_2^{0,3} = 0\) and is involved in a complex of the form \(0 \to E_2^{0,3} \to E_2^{2,2} \to E_2^{4,1} \to E_2^{6,0} \to 0\), which we can identify as \(0 \to 0 \to 0 \to H^4({\mathbf{CP}}^2) \to H^6({\mathbf{CP}}^2) \to 0\), which must be exact, so we have \(H^4({\mathbf{CP}}^2) \cong H^6({\mathbf{CP}}^2)\).

\(E_\infty^{1,2}\): We have the complex \(0 \to E_2^{1,2} \to E_2^{3,1} \to E_2^{5, 0} \to 0\) which equals \(0 \to 0 \to H^3({\mathbf{CP}}^2) \xrightarrow{f} H^5({\mathbf{CP}}^2) \to 0\), which must be exact and so \(f\) is an isomorphism yielding \(H^3({\mathbf{CP}}^2) \cong H^5({\mathbf{CP}}^2)\).

\(E_\infty^{2,1}\): We have the complex \(0 \to E_2^{0, 2} \to E_2^{2, 1} \to E_2^{4, 0} \to 0\) which equals \(0 \to 0 \to H^2({\mathbf{CP}}^2) \to H^4({\mathbf{CP}}^2) \to 0\), so \(H^2({\mathbf{CP}}^2) \cong H^4({\mathbf{CP}}^2)\).

(Here we are using the fact that \(E_2^{0,2} = H^2(S^1) = 0\) instead of the automatic zeros from the differentials extending into zero quadrants.)

\(E_\infty^{3,0}\): We have \(0 \to E_2^{1,1} \to E_2^{3,0} \to 0\) which equals \(0 \to H^1({\mathbf{CP}}^2) \to H^3({\mathbf{CP}}^2) \to 0\) which must be exact and so \(H^1({\mathbf{CP}}^2) \cong H^3({\mathbf{CP}}^2)\)

Note that \(H^4(S^5) = 0\) doesn’t give any new information at this point.

Claim

$$H^5(S^5) = {\mathbf{Z}}\implies H^6({\mathbf{CP}}^2) = 0$$

We have \(H^5(S^5) = \displaystyle\bigoplus_{p+q=n}E_2^{p,q}\), and so there must now be a nonzero term in this sum.

Since \(q > 1\) stabilizes to zero on \(E_2\), the nonzero term must come from \(E_2^{5,0}\) or \(E_2^{4,1}\).

\(E_2^{5,0}\): The complex is \(0 \to H^3({\mathbf{CP}}^2) \to H^5({\mathbf{CP}}^2) \to 0\)

\(E_2^{4,1}\): The complex is \(0 \to H^4({\mathbf{CP}}^2) \to H^6({\mathbf{CP}}^2) \to 0\)

In order for an \(E_3\) term to be nonzero, one of these complexes must have nonzero homology. But by the previous claim, \(0 \to H^3({\mathbf{CP}}^2) \to H^5({\mathbf{CP}}^2) \to 0\) does have zero homology, so we consider the second complex instead.

We know from our current results that \(0 \to H^4({\mathbf{CP}}^2) \to H^6({\mathbf{CP}}^2) \to 0\) is equal to \(0 \to{\mathbf{Z}}\xrightarrow{f} H^6({\mathbf{CP}}^2) \to 0\), and we know that \(\frac{\ker f}{\operatorname{im}0} = \ker f \cong H^5(S^5) = {\mathbf{Z}}\), since this is the only possible nonzero term in the above sum.

(Not sure how to use \(\ker f = 0\) to show \(H^6({\mathbf{CP}}^2) = 0\), or how to inductively compute \(H^*({\mathbf{CP}}^n)\).)