11
$\begingroup$

Let $E$ and $F$ be two complex vector bundles over a space $X$. There's a fairly well-known binary operation called the direct sum, written $E\oplus F$, which has the property that its first Chern class is the sum of the Chern classes of the constituents: $c_{1}(E\oplus F)=c_{1}E+c_{1}F$.

My question is: For $k\geq 1$, are there binary operations $B^{k}(E,F)$ with the property that $c_{k}(B^{k}(E,F))=c_{k}E+c_{k}F$?

I am happy to allow the binary operation to take values in virtual bundles if need be.

Motivation: These operations would of course be pretty useful for contstructing (virtual) vector bundles with desired Chern classes.

$\endgroup$
6
$\begingroup$

Let's work with virtual bundles. Your question is equivalent to the following:

If we fix a $k \geq 1$, does the map $BU \times BU \rightarrow K(\mathbb{Z},2k)$ representing $c_k \otimes 1 + 1 \otimes c_k$ factor through some map $BU \times BU \rightarrow BU$ composed with the map $BU \rightarrow K(\mathbb{Z},2k)$ representing $c_k$. Recall that $BU$ has a CW structure with cells in every even dimension, and that $c_k$ is represented by a single cell. From this description of $c_k$, the second map $BU \rightarrow K(\mathbb{Z},2k)$ can be constructed by extending the map $BU_{2k}/BU_{2k-1} \rightarrow K(\mathbb{Z},2k)$ ($BU_{2k}$ meaning the 2k-skeleton) which on the summand corresponding to $c_k$ represents a generator of $\pi_{2k}(K(\mathbb{Z},2k))$ and elsewhere is constant, to $BU/BU_{2k-1}$ and then precomposing with the quotient $BU \rightarrow BU/BU_{2k-1}$.

Similarly, $c_k \otimes 1 + 1 \otimes c_k$ can be represented in the same manner where the nontrivial maps in the bouquet of spheres correspond to the cells $c_k \otimes 1$ and $1 \otimes c_k$.

For a CW complex X with cells in only even dimensions and a single 0-cell, there is a one to one correspondence between $[X,BU]_*$ and the direct product generated by by the non-basepoint cells of $X$. This is given by letting the coordinate corresponding to the d-cell, which we name e, take $[f]:X \rightarrow BU$ to the class $S^d \xrightarrow{e} X_d/(X_{d-1}-e) \xrightarrow{f} BU$ (see Atiyah's K-theory Proposition 2.5.2).

In this case we consider $BU \times BU/ (BU \times BU)_{2k-1}$, and we choose an element of the product which in the coordinates corresponding to $c_k \otimes 1$ and $1 \otimes c_k$ is represented by the generator of $\pi_{2k}(BU)$. Let's call the corresponding map $\theta$ from $BU \times BU \rightarrow BU \times BU/ (BU \times BU)_{2k-1} \rightarrow BU$. Up to homotopy we then have a factorization of $c_k \otimes 1 + 1 \otimes c_k$ as $c_k \circ \theta$ because the maps $\pi_{2k}(BU \times BU/ (BU \times BU)_{2k-1}) \rightarrow \pi_{2k}(K(\mathbb{Z},2k))$ are the same. This relies on the (I think true) statement that generator of $\pi_{2k}(BU)$ under the composite $\pi_{2k}(BU) \rightarrow\pi_{2k}(BU /BU_{2k-1}) \rightarrow H_{2k}(BU /BU_{2k-1})$ lands on the cell which represents $c_k$.

$\bf{Edit:}$ Unfortunately, I've read that this is untrue. Instead it lands on some multiple of the dual of the Chern class. Denote this multiple $a_k$.

So to apply the operation to the virtual bundles $E,F$ you take the composite $\theta \circ (E,F)$. As well as having the property $c_k(\theta \circ (E,F))=a_k (c_k (E)+c_k(F))$, we also have that $c_l=0$ for $0<l<k$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ For anyone who's interested, $a_{k}=1$ if $k=1,2,4$ and $a_{k}=2$ otherwise, as computed in "Realizing cohomology classes as Euler classes" by A. Naoleka. Always a pleasure to see the Hopf-invariant -1 pattern showing up $\endgroup$ – John Greenwood Jan 8 at 2:43
  • 1
    $\begingroup$ It seems to me that $a_k$ is instead the $k$-th Chern class of the generator of $\pi_{2k}(BU)$, which is $(k-1)!$. This is essentially why the number $6 = (4-1)!$ shows up in my answer. $\endgroup$ – Bertram Arnold Jan 8 at 11:24
  • 1
    $\begingroup$ But after talking with Mike Miller, I think the construction I gave (upon correction) is the following: take your virtual bundles $E,F$over $X$ and use the Bott isomorphism to get virtual bundles over $\Sigma^2 X$. Now over a suspension it is true that $c_k(E \oplus F)=c_k(E)+c_k(F)$, so add these bundles in $\Sigma^2 X$ and then transport them back down to $X$ via the Bott isomorphism. Then the result is that the chern class of this new bundle over $X$ is $a_k(c_k(E)+c_k(F))$ for all $k>0$. And this $a_k=(k-1)!$ $\endgroup$ – Connor Malin Jan 8 at 23:24
  • 1
    $\begingroup$ @ConnorMalin I thought so :). I think i figured out why the $c_{k}$ of the generator is $(k-1)!$ - it's given by the $k-1$-fold (external) tensor product with the Bott class $[L]-1$ in $\tilde{K}(S^2)$. The you plug and chug with the rules for chern classes of direct sums and tensor products of line bundles. $\endgroup$ – John Greenwood Jan 9 at 1:19
  • 1
    $\begingroup$ The factor shows up since the $k$-th component of the Chern character is $(-1)^{k+1}\frac{c_k}{(k-1)!} + $polynomial in $c_1,\dots,c_{k-1}$. The factor is $\frac{1}{k!}$ times the coefficient of $x_1\dots x_n$ in the expression of $x_1^k+\dots+x_k^k$ in terms of symmetric polynomials, which is $(-1)^{k+1} k$ (compare en.wikipedia.org/wiki/… ). Thus $\frac{c_k((L-1)^{\boxtimes k})}{(k-1)!}$ generates the image of the Chern character in $H^{2n}(S^{2n};\mathbb Q)$ which by multiplicativity is $H^{2n}(S^{2n};\mathbb Z)$. $\endgroup$ – Bertram Arnold Jan 9 at 9:24
8
$\begingroup$

There is such an operation for $k = 2$ using virtual bundles.

Note that $c_2(E\oplus F) = c_2(E) + c_1(E)c_1(F) + c_2(F)$ so

\begin{align*} c_2(E) + c_2(F) &= c_2(E\oplus F) - c_1(E)c_1(F)\\ &= c_2(E\oplus F) - c_1(\det E)c_1(\det F)\\ &= c_2(E\oplus F) - c_2(\det E \oplus \det F). \end{align*}

Recall that the second Chern class of a virtual vector bundle $A - B$ is given by

$$c_2(A - B) = c_2(A) - c_2(B) - c_1(A)c_1(B) + c_1(B)^2$$

which reduces to $c_2(A) - c_2(B)$ if $c_1(A) = c_1(B)$. As

$$c_1(E\oplus F) = c_1(E) + c_1(F) = c_1(\det E) + c_1(\det F) = c_1(\det E\oplus\det F),$$

the above computation shows

$$c_2(E) + c_2(F) = c_2(E\oplus F - \det E\oplus\det F).$$ That is, we can take $B^2(E, F) = E\oplus F - \det E\oplus \det F$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, this is wonderfully concrete and easy to apply in practice! $\endgroup$ – John Greenwood Jan 8 at 18:26
5
$\begingroup$

Such an operation with values in bundles does not exist for $k = 4$ and the base space $\mathbb{HP}^2$. For virtual vector bundles, it depends how you extend the definition of the Chern classes; for the natural extension such that the total Chern class is multiplicative under sums, it also doesn't exist (in fact, this extension is used in the proof).

Pullback along the evident inclusion $\mathbb{HP}^n\hookrightarrow \mathbb{HP}^\infty\cong BSU(2)$ gives a ring homomorphism $R(SU(2))\to K^0(\mathbb{HP}^n)$ preserving the augmentation ideals $I$. Since $\mathbb{HP}^n$ is covered by $n+1$ contractible sets (the standard coordinate charts), we have $I^3_{K^0(\mathbb{HP}^n)} = 0$, and therefore an induced morphism $\phi_n:R(SU(2))/I_{R(SU(2))}^{n+1}\to K^0(\mathbb{HP}^n)$. Now the representation ring $R(SU(2))$ is additively generated by the representations $[V_n]$, of dimension $n+1$, subject to the relations $[V_m]\otimes [V_n] = [V_{|m-n|}] + [V_{|m-n|+2}] + \dots + [V_{m+n}]$. Setting $v = [V_1]-2$, it follows easily that $I_{R(SU(2))}^n/I_{R(SU(2))}^{n+1}$ is generated by $v^n$; multiplicativity of the Atiyah-Hirzebruch spectral sequence then shows that $\phi_n$ is an isomorphism. (This calculation is a special case of the Atiyah-Segal completion theorem.)

Now $I_{R(SU(2))}/I_{R(SU(2))}^3$ is generated by the classes of $V_1-2$ and $V_2-3$, and their images $E_1-2$ and $E_2-3$ define virtual vector bundles over $\mathbb{HP}^2$ whose classes generate $\widetilde K^0(\mathbb{HP}^2)$. Let's calculate their total Chern classes: $E_1$ is just the canonical $2$-dimensional bundle over $\mathbb{HP}^2$, so its total Chern class is $c(E_1) = 1 + x$ where $x\in H^4(\mathbb{HP}^2;\mathbb Z)$ is a generator. Since $E_2$ is $3$-dimensional, only its second Chern class can be nonzero, and one easily calculates $c(E_2) = 1 + c_2(E_2) = 1 + 4x$: we have $E_2\oplus \mathbb C = E_1^{\otimes 2}$, so pulling back along the map $\iota:\mathbb{CP}^2\to \mathbb{HP}^2$, noting that $\iota^*E_{1} = L\oplus L^{-1}$ splits as a sum of the canonical line bundle and its dual and setting $u = c_1(L)$ we have \begin{align*} \iota^*c(E) &= (1+u)(1-u) = 1-u^2\\ \implies \iota^*x &= -u^2\implies \iota^*|_{H^4}\text{ is injective}\\[12pt] \iota^*E_2 &= (\iota^*E_1)^{\otimes 2} - 1 \\ &= (L\oplus L^{-1})^{\otimes 2} - 1 \\ &= L^{\otimes 2} + L^{\otimes -2} + 1\\ \iota^*c(E_2) &= (1 + 2u)(1 - 2u)\\ &= 1 - 4u^2 \end{align*}

The total Chern class can be extended to a natural transformation $K^0(X) \to \bigoplus H^{2*}(X;\mathbb Z)$ by $[E] - [F]\mapsto c(E)c(F)^{-1}$. We immediately obtain \begin{align*} c(k\cdot E_1 + l\cdot E_2) &= (1+x)^k(1+4x)^l \\ &= 1 + (k + 4l)x + \left(\frac{k(k-1)}{2} + 4kl + 8l(l-1)\right)x^2\\ &= 1 + (k+4l)x + \left(\frac{(k+4l)(k+4l-1)}{2} - 6l\right)x^2 \end{align*}

In particular, no cohomology class of the form $(3n+2)x^2$ with $n\in\mathbb Z$ can be the fourth Chern class of a (virtual) bundle over $X$, so there is no (virtual) bundle $F$ with $c_4(F) = c_4(E_1^{\oplus 2}) + c_4(E_1^{\oplus 2})$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, this is nice! I took the liberty to change some minor typos. $\endgroup$ – John Greenwood Jan 8 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.