A note on torsion in K-theory – That Can't Be Right

In this post we will look at the correspondence between torsion in cohomology of finite CW-complexes and torsion in their K-theory. The title could just as well have been “a use case for Atiyah-Hirzebruch spectral sequence” as this will be our primary tool.

Throughout this post $X$ will denote a finite CW-complex, though much of the discussion carry over to more general spaces with finitely generated cohomology groups. Note that $X$ is automatically a compact and Hausdorff space, but not all compact Hausdorff spaces are homotopy equivalent to a CW-complex.

We will be looking at torsion in cohomology and how it relates to torsion in K-theory of the space $X.$ Torsion in cohomology triggers various anxiety symptoms in many geometers, that often results in a quick tensoring with $\R$ . Unfortunately torsion is at times very important to keep track of, and the lack of tools to handle this situation cannot justify ignoring it.

Examples of spaces with torsion in cohomology are the projective spaces which we will look at below in Example 1 or more generally Lens spaces. Other families of examples can be constructed from classifying spaces of discrete groups $\Gamma$ finite abelianization (for example a group with Kazhdan’s property T) many of which admit a classifying space with the homotopy type of a finite CW-complex (by the so called Borel-Serre compactification of the classifying space).

The Atiyah-Hirzebruch spectral sequence (henceforth the AHSS) gives a way to calculate the K-theory (or other generalized (co)homology theories) of $X$ by solving a collection of exact sequences involving the cohomology of $X$ . The $E_2$ -page of the sequence looks like

$E_2^{p,q} = H^p(X, K^q(\star))$

where $\star$ is the one point space.

To calculate the K-theory from the spectral sequence, it is assumed that a filtration of $K^*(X)$ is given corresponding to the spectral sequence. Since we are dealing with a CW-complex, there is a standard way to construct such a filtration using the CW-structure of $X$ as follows:

$X_0\subset X_1 \subset ... \subset X_n$

is the filtration of $X$ determined by the CW-structure, where $dim(X_i) = i$ ,

$F^{n}K^i(X):= ker[K^i(X) \to K^i(X_n)]$

and note that this gives a filtration

$0 = F^{n+1}K^i(X) \subset F^{n}K^i(X) \subset F^{n-1}K^i(X) ... \subset F^0K^i(X) = K^i(X)$

of the $i$ ‘th K-group $K^i(X)$ . With the knowledge of the so called $E_\infty$ -page of the AHSS, we can (in theory) calculate the K-theory of $X$ using the short exact sequences

( $\star$ ) $0\to F^{s+1}K^i(X) \to F^sK^i(X) \to E^{s,s+i}_\infty \to 0$

This is easy in the case $Ext(E^{p,q}_\infty, F^{p+1}K^q(X)) = 0$ for all $p,q$ , i.e. if for instance $E^{p,q}_\infty$ are all free $\Z$ -modules or $\Q$ -vector spaces, since then all the extensions of equation ( $\star$ ) split as

$F^sK^i(X) = F^{s+1}K^i(X) \oplus E^{s,s+i}_\infty$

and the resulting $K$ -theory groups are simply given by the diagonal direct sums

$K^i(X) = \bigoplus_{t=0}^\infty E^{t, t+i}_\infty.$

In general this is not so easy and there are examples where there is torsion on the $E_\infty$ -page, but the resulting $K$ -theory groups are torsion free (see for instance this article).

Before going further down the rabbit hole, let us quickly look at what happens when either $K^*(X)$ or $H^*(X)$ is know to be torsion free:

The torsion free case

There is another way to relate the K-theory of the space $X$ with the cohomology of the space $X$ , namely using the Chern maps:

$\begin{align*} Ch: & K^0(X) \to \bigoplus_{i=0}^\infty H^{2i}(X, \Z) \otimes \Q \\ Ch: & K^1(X) \to \bigoplus_{i=0}^\infty H^{2i-1}(X, \Z) \otimes \Q. \end{align* }$

These maps are functorial and induce induce maps of AHSS –

$E_2^{p,q} = H^p(X, K^q(\star)) \to \hat{E}_2^{p,q} = H^p(X, H^q(\star, \Q))$

of AHSS associated with the K-theory and rational cohomology theory of $X$ . The left hand side spectral sequence is known to collapse at the $E_2$ -page, from which Atiyha and Hirzebruch were able conclude (see this article) that the Chern maps are rational isomorphisms, meaning

$\begin{align*}Ch\otimes id :& K^0(X)\otimes \Q \to \bigoplus_{i=0}^\infty H^{2i}(X, \Z) \otimes \Q \\Ch\otimes id :& K^1(X)\otimes \Q \to \bigoplus_{i=0}^\infty H^{2i-1}(X, \Z) \otimes \Q\end{align* }$

are both isomorphisms. In particular $Ker(Ch) \subset K^i(X)$ are torsion subgroups. An immediate consequence is the following proposition:

Proposition 1(Atiyah, Hirzebruch) If $K^0(X)$ is torsion free, then $Ch$ is injective (similarly for $K^1$ ). Thus we have $K^0(X) = \Z^{b^{even}}$ (resp. $K^1(X) = \Z^{b^{odd}}$ ) is a free abelian group with the rank $b^{even} = \sum_{i= 0}^\infty b_{2i}$ (resp. $b^{odd}: = \sum_{i= 0}^\infty b_{2i-1}$ ) where $b_i$ are the Betti numbers of $H^*(X, \Z)$ .

Two observations should be made. First we stress that this is an isomorphism of abstract $\Z$ -modules. There is no reason to expect the image of $Ch(K^0) \subset H^{even}(X,\Z) \subset H^{even}(X, \Z)\otimes \Q$ to land in the integral cohomology, and even if it did, it could very well have a non-trivial cokernel. Second, $H^*(X)$ may still have torsion in the above Proposition. If however we know $H^*(X)$ has no torsion we get the following:

Proposition 2 (Atiyah, Hirzebruch) If $H^*(X, \Z)$ is torsion free, then the Chern map is injective, thus in this case we have an isomorphism of $\Z$ -modules

$K^0(X) = \bigoplus_{i=0}^\infty H^{2i}(X, \Z)$

and

$K^1(X) = \bigoplus_{i=0}^\infty H^{2i-1}(X, \Z).$

Let us state one more important case where the K-theory turn out to be torsion free. The following is taken from the book of Hatcher (Vector bundles and K-theory) Prop. 2.23 or Proposition 2. 5. 2 of Atiyah’s K-theory

Property 3: Assume $X$ consists of $n$ cells, all of which have even dimension. Then

$K^0(X) = \Z^n \quad \text{and} \quad {K^1(X) = 0.$

Examples of such spaces are the complex projective space or more generally complex Grassmannians, or (more general still) generalized complex flag manifolds given as quotients $G/P$ of a connected complex semisimple Lie group $G$ with finite center over a minimal parabolic subgroup $P$ . There is an elegant way to compute the K-theory of these genralized flag manifolds using the root data of the group $G$ and a K-theoretic equivalent of the Schubert calculus of cohomology theory. See this article for more.

A common folklore is that “the Chern map is an integral isomorphism in lower dimensions”. If we interpret this as meaning the image of the Chern map lies in $H^*(X, \Z)$ , the folklore is not quite true (see for instance this mathoverflow answer). The statement means that the lattice of points reached by the Chern maps are isomorphic to $H^{even}(X,\Z)$ and $H^{odd}(X, Z)$ as abstract $\Z$ -modules. Let us write it in the form of a proposition which we will generalize later in Proposition 10 –

Property 5: If $X$ is of dimension $\leq 3$ , then

$K^0(X) = \bigoplus_{i=0}^\infty H^{2i}(X, \Z)$

and

$K^1(X) = \bigoplus_{i=0}^\infty H^{2i-1}(X, \Z)$

Note that in case $X$ is oriented and of dimension $\leq 3$ , $X$ has torsion free cohomology which can be deduced from the universal coefficient theorem, so Proposition 5 would in this case follow from Proposition 2.

Now that we have seen some cases where there is no torsion in K-theory, let us move on to more interesting stuff –

Torsion!!

So what can be said about torsion in $K^*(X)$ -theory and how does it relates to torsion in $H^*(X)$ ?

A general rule of thumb is that K-theory has “less” torsion than cohomology theory. We will now make this statement precise by going through several interpretations of what it could mean. Our guide in this section will be the the exercise on p. 520 of Fomenko, Fuchs, Homotopical Topology, which reads:

Prove that the order of the torsion in the groups ,  divides
the order of the torsion in the groups , , respectively.

Let us try to decipher what this exercise could mean

False Proposition – The order of a torsion element, i.e. the size of the subgroup it generates, in $K$ -theory divides, the order of some torsion element in cohomology.

One may think this should be true from the fact that the torsion part of the $E_\infty$ -page of the AHSS consists of elements whose order indeed divides the order of some torsion element in $H^*(X, \Z)$ . This clearly does not imply $K^*(X)$ shares this property as it is determined inductively by solving several short exact sequences. For instance for the extension

$0\to \Z_2\to \Z_6 \to \Z_3 \to 0$

we see that the middle term can easily have higher order than either of the other terms. Let us give an explicit example contradiction the above proposition (see this article for more) –

Example 1 (Karoubi, Corollary 6.47 ) For the real projective spaces, we have

$K^0(\R P^n) = \Z\oplus \Z_{2^{n/2}} \qquad K^1(\R P^n) = \Z$

if $n$ is even or

$K^0(\R P^n) = \Z\oplus \Z_{2^{(n-1)/2}} \qquad K^1(\R P^n) = \Z$

if $n$ is odd. In particular for $n = 4$ we have

$H^{even}(\R P^4) = \Z\oplus \Z_2^2 \neq \Z\oplus \Z_4 = K^0(\R P^4)$

and we see there is no element in $H^{even}(\R P^4)$ whose order is divisible by 4.

However one should note that if

$0\to A \to C \to B \to 0$

is a short exact sequence of finitely generated abelian group and if $C$ has torsion $p$ where $p$ is a prime, then either $A$ or $B$ has a torsion element of order $p$ . In particular we get (see also this article)

Proposition 6 If $K^0(X)$ (resp. $K^1(X)$ ) has p-torsion, where $p$ is a prime, then so does $H^{even}(X)$ (resp. $H^{odd}(X)$ ).

This is most likely what the authors meant by the exercise, but for sake of argument let us state some more false propositions –

False Proposition: The order of a torsion element in $H^{even}(X)$ or $H^{odd}(X)$ divides the order of some torsion element in $K^0(X)$ or $K^1(X)$ respectively.

This proposition is false for the silly reason that torsion in $H^*(X)$ does not imply there is torsion in $K^*(X)$ . Torsion elements in $H^*(X)$ can simply disappear. Several examples of this phenomenon can be constructed using the fundamental theorem of Hodgkin. The group $Spin(n)$ is one such example for $n \geq 7$ (see Mimura and Toda Topology of Lie groups II, Theorem 16 and the subsequent list of examples therein).

Theorem 7 (Hodgkin, Theorem 1.1 ) Let G be a connected, semi-simple and simply connected Lie group then $K^*(G)$ is torsion-free.

Looking again at Example 1, we see that though the group $K^0(\R P^4)$ has elements which generate torsion subgroups of higher order than any element in $H^{even}(\R P^4)$ the size (= order) of the torsion subgroup (i.e. the subgroup of all torsion elements) of $K^0(\R P^4)$ is $4$ and equal to that of $H^{even}(\R P^4)$ . This makes for another interpretation of the exercise, which we summarize in the following proposition –

False Proposition The order of the torsion subgroup of $K^0(X)$ and $K^1(X)$ (assuming they are non-zero) divides the order of the torsion subgroup in $H^{even}(X)$ and $H^{odd}(X)$ respectively.

At least this does not contradict any of the previous examples. However, as the next example shows (taken from here) it is still very much wrong. Other examples can be constructed using Lens spaces –

Example 2: Let $X = \R P^3 \times \R P^5$ . Then

$K^0(X) = \Z^2 \oplus \Z_4\oplus \Z_2^2 \qquad K^1(X) = \Z^2 \oplus \Z_4\oplus \Z_2^2$

while

$H^{even}(X)= \Z^2 \oplus \Z_2^5 \qquad H^{odd}(X) = \Z^2 \oplus \Z_2^5$

The order of the torsion subgroups of $K^0(X)$ and $K^1(X)$ is $8$ while the order of the torsion subgroups of $H^{even}(X)$ and $H^{odd}(X)$ is $10$ , neither of which divides the other. But they seem somehow related. For one, the order of the torsion subgroup in $K$ -theory is smaller than the order of the torsion subgroups in cohomology and has fewer generators. We also have $10 - 8 = 2$ and $2$ divides $10$ . We could venture a guess

Proposition 8 The order of the torsion subgroup of $K^0(X)$ and $K^1(X)$ is less than or equal to the order of the torsion subgroup in $H^{even}(X)$ and $H^{odd}(X)$ respectively.

Let us prove the above proposition. First note that for a short exact sequence of finitely generated abelian groups

$0\xrightarrow{\iota} A \to C\xrightarrow{p} B \to 0$

$0\to tor(A) \xrightarrow{\hat{\iota}} tor(C) \xrightarrow{\hat{p}} tor(B)$

where the first map is injective but second map may fail to be surjective. Thus

$|tor(A)| + |tor(B)| = |ker(\hat{p})| + |tor(B)| \geq |ker(\hat{p})| + |im(\hat{p})| = |tor(C)|.$

From these observations, using AHSS we can deduce the the order of the torsion subgroup of $K^0(X)$ and $K^1(X)$ is less than or equal to the order of the torsion subgroups in

$\bigoplus_{p = 0}^\infty E^{p,p}_\infty \quad \text{and } \quad \bigoplus_{p = 0}^\infty E^{p,p+1}_\infty$

respectively. Now successive applications of the third isomorphism theorem implies there are subgroups $I^{p,q} \subset K^{p,q} \subset H^p(X, K^q(\star))$ such that $H^p(X, K^q(\star))/ K^{p,q}$ and $I^{p,q}$ are finite groups and

$E^{p,q}_\infty = K^{p,q}/ I^{p,q}$

hence $|tor(E^{p,q}_\infty)| \leq |tor(H^p(X, K^q(\star)))|$ from which the proposition readily follows.

So the torsion subgroup of $K^*(X)$ is bounded in size by the size of the torsion subgroup of $H^*(X)$ . But there is more! We can say something about the total number of generators of the torsion subgroup of $K^*(X)$ :

Property 9: The (minimal) number of generators of $H^{even}(X)$ and $H^{odd}(X)$ are greater or equal to the (minimal) number of generators for $K^0(X)$ and $K^1(X)$ respectively.

This is proved in a similar way to the preceding proposition: First note that every element $E_\infty^{p,q}$ is a quotient of subgroups $I^{p,q} \subset K^{p,q} \subset H^p K^q(\star)}.$ It follows that $E_\infty^{p,q}$ have at most the same number of generators as $H^p(X, K^q(\star))$ . Next note that if $A$ and $B$ are abelian groups with generators $(a_1,..,a_n)$ and $(b_1, ..., b_m)$ respectively, then if

$0\to A\to C\to B\to 0$

is an extension, it is easy to check that $C$ can be generated by $n+m$ generators (maybe less). Thus the minimal number of generators of the $K^0(X)$ and $K^1(X)$ is bounded by the number of generators of the diagonal sums

$\bigoplus E_\infty^{p, p} \quad \text{and } \quad\bigoplus E_\infty^{p, p+1}$

respectively, which in turn are are less than or equal to the number of generators $H^{even}(X)$ and $H^{odd}(X)$ respectively.

We have mentioned the rational Chern maps are isomorphisms so we know that the generators of the free abelian subgroups of $H^{even}(X)$ and $H^{odd}(X)$ are in 1-1 correspondence with the generators of the free abelian subgroups of $K^0(X)$ and $K^1(X)$ respectively. It is the number of generators of the torsion subgroups that could be smaller in $K$ -theory than cohomology.

How much torsion is in K-theory

The preceding section showed, in addition to the importance of writing solutions to ones exercises, that the size of the torsion subgroups and the number of its generators in $K$ -theory are bounded by those in cohomology. In some situations it is possible to bound these two quantities from below as well.

The main Proposition we will mention in this regard is the following –

Property 10 (source): Assume $X$ has dimension $\leq 2n + 1$ (resp. $\leq 2n$ ), then the assignment which sends a vector bundle $E\to X$ to its (total) Chern class

$K^0(X)\ni [E] \mapsto c(E) = c_0(E) + c_2(E) + ... \in H^{even} (X)$

determines an map

$K^0(X) \to H^{even}(X) \quad (\text{resp. } K^1(X) \to H^{odd}(X))$

which is an isomorphism up to $p$ -torsion, where $p$ runs over all primes $p< n$ .

In particular if $H^{even}(X)$ has a single torsion generator of prime order $p \geq (dim(X)-1)/2$ then Proposition 10 above implies

$K^0(X) = H^{even}(X).$

Proposition 10 could probably be deduced from the known bounds on the exponential of the differentials $\partial^{p,q}_n : E^{p,q}_n \to E^{p-n, q+n-1}$ of the AHSS computed by D. Arlettaz here. However the proof given by Thomas in Theorem 1 of A relation between k-theory and cohomology is so elegant that I repeat it here

Proof (sketch) The assignment of a vector bundle on $X$ to its total Chern class $c(X) = c_0(X) + c_1(X) + ... \in H^*(X)$ determines a natural transformation of half-exact functors

$K^0 \to H^{even}$

Now by a theorem of Dold (see Proposition 2.1 of these notes) if $W$ is the category whose objects are finite CW-complexes and morphisms are homotopy classes of maps. Then any two half exact functors (topologically half exact would suffice here) $t$ and $t'$ from $W$ to the category of abelian groups that agree on the spheres $S^i$ for $i\leq m$ , modulo some fixed Serre class of abelian groups, must agree for all $X\in W$ of dimension $dim(X)\leq m$ modulo the Serre class.

Atiyah and Hirzebruch have shown that the map

$K^0(S^n) \to H^{even}(S^n)$

are injective with cokernel given by $\Z_{(n-1)!}$ .

Thus it is an isomorphism modulo a $p$ -torsion, where $p$ runs over all primes in $p< n$ . The claim for $K^1$ and $H^{odd}$ follows by replacing $X$ with its suspension $SX$ , though this has one dimension more than $X$ , so we need to adjust the conditions on the dimension of $X$ accordingly.

Note how Proposition 10 implies proposition 5, since if $dim(X) \leq 3$ we have no primes strictly less than $2$ so the map in Proposition 10 is an isomorphism.

The torsion free case

Torsion!!

How much torsion is in K-theory

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