---- The Kunneth theorem for [[singular cohomology]] relates the [[singular cohomology|cohomology]] of a [[product topology|product space]] to that of the factors. > [!theorem] Theorem. ([[Kunneth Theorem for singular cohomology]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]], $Y$ a [[topological space]] such that $H^{n}(Y;R)$ is a [[free module|free]] $R$-[[module]] for each $n.$ Then the [[cross product on cohomology|cross product map]] $\bigoplus_{k+\ell=n}H^{k}(X;R) \otimes H^{\ell}(Y;R) \xrightarrow{\times}H^{n}(X \times Y;R)$ > is an [[isomorphism]] for every $n$, for every finite [[cell complex]] $X$. > > For convenience, one often writes $H^{*}(X;R) \otimes H^{*}(Y;R)$ for the [[graded module|graded]] $R$-[[module]] which in grade $n$ is given by $\bigoplus_{k+\ell=n}H^{k}(X; R) \otimes H^{\ell}(Y;R).$ > Then Kunneth says the map given by $H^{*}(X ;R) \otimes H^{*}(Y;R) \xrightarrow{\times} H^{*}(X \times Y;R)$ > is an [[isomorphism]] of [[graded ring|graded rings]] (in fact, of graded $R$-[[algebra|algebras]]). > [!proof]- Proof. ([[Kunneth Theorem for singular cohomology]]) > ~ > [!note] Remark. The conditions of the theorem require $H^{n}(Y;R)$ to be free. This holds e.g. if $R=\mathbb{F}$ is a [[field]], then all [[module|modules]] are free. So $H^{*}(X \times Y; \mathbb{F}) \cong H^{*}(X;\mathbb{F}) \otimes H^{*}(Y; \mathbb{F})$ for any [[field]] $\mathbb{F}$. > [!basicexample] Example. (Cohomology ring of the torus) > Consider $H^{*}(\mathbb{S}^{1}; \mathbb{Z})$. We know $H^{i}(\mathbb{S}^{1};\mathbb{Z})=\begin{cases} > \mathbb{Z}\langle 1 \rangle & i=0 \\ > \mathbb{Z}\langle x \rangle & i=1 \\ > 0 & i \geq 2, > \end{cases}$ > > where $1$ is the [[singular cohomology|identity class]]. > > For the generator $x$ of $\mathbb{Z}\langle x \rangle$, have $x \smile x=0$ because there's nothing in degree $2$. So we know $H^{*}(\mathbb{S}^{1}; \mathbb{Z})= \frac{\mathbb{Z}[x]}{\langle x^{2} \rangle }.$ > Then Kunneth's theorem tells us that $H^{*}(T^{n}; \mathbb{Z}) \cong H^{*}(\mathbb{S}^{1}; \mathbb{Z})^{\otimes n}$ > where $T^{n}=(\mathbb{S}^{1})^{n}$ is the $n$-torus and this is a [[ring isomorphism]]. This is the [[exterior algebra]] in $n$ generators $\Lambda^{n}[x_{1},\dots,x_{n}]=\frac{\mathbb{Z}[x_{1},\dots,x_{n}]}{\langle x^{2}_{i}, x_{i}x_{j}+x_{j}x_{i} \rangle }$ > using the fact that $x_{i}, x_{j}$ have degree $1$ so anticommute. Warning about noncommutativity of $\mathbb{Z}[x_{1},..,x_{n}]$ as written here. > [!basicexample] > Let $f:\mathbb{S}^{n} \to T^{n}$ be a map for $n \geq 2$. Claim: the induced morhpism in top-degree $f^{*}:\underbrace{ H^{n}(T^{n};\mathbb{Z}) }_{ \cong \mathbb{Z} } \to \underbrace{ H^{n}(\mathbb{S}^{n}; \mathbb{Z}) }_{ \cong \mathbb{Z} }$ > is zero. Indeed, looking at the [[presentation of a group|presentation]] above, we know that $H^{n}(T^{n}; \mathbb{Z})$ is generated by $x_{1} \smile \dots \smile x_{n}$.[^1] Pulling back (recall $f^{*}$ is a [[ring homomorphism]]), $f^{*}$ sends this to $f^{*}(x_{1}) \smile \dots \smile f^{*}(x_{n})$. But $f^{*}(x_{i}) \in H^{1}(\mathbb{S}^{n}; \mathbb{Z})=0$ must be trivial for all $n \geq 2$. So $f^{*}$ kills the sole generator of $H^{n}(T^{n};\mathbb{Z})$, thus kills everything. > > This shows the importance of Kunneth: it tells us not only the cohomology, but the ring structure on it. ---- #### [[cup product]] [[singular cohomology]] [[(co)homology of a complex|cohomology]] [[tensor product of modules|tensor product]] $H^{*}(Y;R) \otimes H^{*}(X;A) \xrightarrow{\times} H^{*}()$ [^1]: Think of as the generator of top-degree nonvanishing [[differential form]]s. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```