group cohomology, nonabelian group cohomology, Lie group cohomology
cohomology with constant coefficients / with a local system of coefficients
differential cohomology
(also nonabelian homological algebra)
A Künneth theorem is a statement relating the homology or cohomology of two objects $X$ and $Y$ with that of their product $X \times Y$.
In good situations it identifies the (co)homology of a product with the tensor product of the (co)homologies: $H^\bullet(X \times Y, E) \simeq H^\bullet(X,E) \otimes H^\bullet(Y,E)$. But in general this simple relation receives corrections by Tor-groups.
We discuss the Künneth theorem in ordinary homology.
Let $R$ be a ring and write $\mathcal{A} = R$Mod for its category of modules. (For instance $R = \mathbb{Z}$ the integers, in which case $\mathbb{Z} Mod \simeq$ Ab is the category of abelian groups. )
For $N_1, N_2 \in R Mod$ two modules, write $N_1 \otimes N_2 \in R Mod$ for their tensor product of modules. Similarly for $C_\bullet, C'_\bullet \in Ch_\bullet(R Mod)$ two chain complexes of $R$-modules, write $(C \otimes_R C')_\bullet$ for their tensor product of chain complexes. Finally write $Tor^R_1(N_1,N_2)$ for the first Tor-module of $N_1$ with $N_2$.
We discuss the Künneth theorem over $R$ in stages, starting with important special cases and then passing to more general statements.
All these versions hold for chain homology and tensor products of general chain complexes. But under the Eilenberg-Zilber theorem all these statements apply directly in particular to the singular homology of topological spaces and their products. This is discussed below in
Let $R = k$ be a field.
For $R = k$ a field, given two chain complexes of $k$-vector spaces $C_\bullet,C'_\bullet \in Ch_\bullet(k Vect)$, for each $n \in \mathbb{N}$ there is an isomorphism
between the chain homology of the tensor product of chain complexes and the tensor product of abelian groups of chain homologies.
For a proof see the proof of theorem below, of which this is a special case.
Let now $R$ be a ring which is a principal ideal domain. This may be a field as above, or for instance it may be the ring $R = \mathbb{Z}$ of integers, in which case $R$-modules are equivalently just abelian groups.
For $R$ a principal ideal domain, given a chain complex $C_\bullet \in Ch_\bullet(R Mod)$ of free modules over $R$ and given any other chain complex $C'_\bullet \in Ch_\bullet(R Mod)$, then for each $n \in \mathbb{N}$ there is a short exact sequence of the form
This appears for instance as (Hatcher, theorem 3B.5).
In the special case that $C'$ is concentrated in degree 0, this is the universal coefficient theorem in ordinary homology.
In particular if all the Tor-groups on the right vanish, then the theorem asserts an isomorphism
This is the case (assuming the axiom of choice) notably if $R$ is a field (since every module over a field is a free module – every vector space has a basis – and every free module is a flat module).
Notice that since $C_k$ is assumed to be free, hence a direct sum of $R$ with itself, since the tensor product of modules distributes over direct sums, and since chain homology respects direct sums, we have
First consider now the special case that all the differentials of $C_\bullet$ are zero, so that $H_k(C_\bullet) = C_k$. In this case (1) yields $H_n(C_k \otimes_R C') \simeq H_k(C_\bullet) \otimes_R H_{n-k}(C'_\bullet)$ and therefore
Since $H_k(C) = C_k$ is a free module by assumption, it has no Tor-terms (by the discussion there) and hence this is the statement to be shown.
Now let $C_\bullet$ be a general chain complex of free modules. Notice that for each $n$ the cycle-chain-boundary-short exact sequence
splits due to the assumption that $C_n$ is a free module, and hence (as discussed at split exact sequence) that it exhibits a direct sum decomposition $C_n \simeq Z_n \oplus B_{n-1}$. Since the tensor product of modules distributes over direct sum, it follows that tensoring with any $C'_k$ yields another short exact sequence
This means that if we regard the graded modules $Z_\bullet$ and $B_\bullet$ of chains and of boundaries as chain complexes with zero-differentials, then we have a short exact sequence of chain complexes
This induces its homology long exact sequence of the form
Here the terms involving the complexes $B$ and $Z$ of boundaries and cycles may be evalutated, since these have zero differentials, via the special case discussed at the beginning of this proof to yield the long exact sequence
where $i_n \coloneqq H_n(i \otimes C')$ is the morphism induced from the inclusion $i \colon B_\bullet \hookrightarrow Z_\bullet$ of boundaries into cycles.
This means that by quotienting out an image on the left and a kernel on the right, we obtain a short exact sequence
Since the tensor product of modules is a right exact functor it commutes with the cokernel on the left, as does the formation of direct sums, and so we have
This is the left term in the short exact sequence to be shown. For the right term the analogous argument does not quite go through, because tensoring is not in addition a left exact functor, in general. The failure to be so is precisely measured by the Tor-module:
Notice that by the assumption that $C_n$ is free and using the fact (discussed at principal ideal domain) that over our $R$ the submodules $B_n, Z_n \hookrightarrow C_n$ are themselves free modules, the defining short exact sequence $0 \to B_n \stackrel{i_n}{\to} Z_n \to H_n(C) \to 0$ exhibits a projective resolution of $H_n(C)$. Therefore by definition of Tor we have
This identifies the term on the right of the exact sequence to be shown.
For $R$ any ring, there is a spectral sequence converging to the homology of the tensor product, whose second page involves all the Tor-modules: the Künneth spectral sequence
(…)
For instance (Williams, section 4.2).
Let $R$ be a ring and let $X,Y$ be topological spaces. If the ordinary cohomology ring $H^k(Y,R)$ is a finitely generated free module over $R$, then the comparison map (“cross product”)
(from the tensor product of cohomology rings to the cohomology ring of the product space) is an isomorphism.
(e.g. Hatcher, theorem 3.15, and in more generality: Spanier, section 5.5, theorem 11)
The Künneth theorem for generalised cohomology theories is a special case of the universal coefficient theorem. Let $E$ be a ring spectrum, $X$ and $Y$ two spectra. Then we define a new cohomology theory as the $E$-module spectrum $G \coloneqq F(Y, E)$ where $F(-,-)$ is the function spectrum (that is, the internal hom in the category of spectra). The (reduced) cohomology of $X$ in this cohomology theory is thus given by:
A universal coefficient theorem gives a way of computing $G^*(X)$ from knowledge of $E^*(X)$ and $G^*(pt)$. In this case, $G^*(pt) = E^*(Y)$ so our initial data is $E^*(X)$ and $E^*(Y)$.
For more details, see the page on the universal coefficient theorem.
All of these statements have their analogs for singular homology of topological spaces $X , Y \in$ Top: by the Eilenberg-Zilber theorem there is a quasi-isomorphism
between the singular chain complex of the product space and the tensor product of chain complexes of the separate singular chain complexes. Hence in particular there are isomorphisms of singular homology
Using this in the above statements of the Künneth theorem yields directly the Künneth theorem for singular homology of topological spaces.
The original articles are
H. Künneth, Über die Bettischen Zahlen einer Produktmannigfaltigkeit Math. Ann. , 90 (1923) pp. 65–85
H. Künneth, Über die Torsionszahlen von Produktmannigfaltigkeiten Math. Ann. , 91 (1924) pp. 125–134
Textbook accounts include
Lecture notes include
Section 3.B of
Section 3.6
Section 4.2 in
See also the corresponding references at universal coefficient theorem.
In the context of parameterized spectra
Last revised on October 7, 2020 at 13:47:42. See the history of this page for a list of all contributions to it.