Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could anyone point me to a reference where the formal group law of (topological or motivic) K-theory is computed in as much detail as possible?

share|cite|improve this question

Well, I don't think there can be much detail here: the first Chern class of a line bundle in topological K-theory is defined by $c_1(L)=\pm1\pm[L]$; hence $c_1(L\otimes L')=c_1(L)+c_1(L')\pm c_1(L)c_1(L')$ — so the FGL is just $F(u,v)=u+v\pm uv$ (sign convention varies).

share|cite|improve this answer

Let $Vect^1(X)$ be isomorphism classes of complex line bundles on $X$. I assume you're familiar with the definition of topological $K^0(X)$.

Let's examine the map $c_1: Vect^1(X) \to K^2(X)$, (recalling that, due to Bott periodicity, $K^0(X) \simeq K^2(X)$). The group operators are the tensor product of line bundles (in $Vect^1(X)$) and the tensor product of virtual line bundles (in $K^0(X)$).

Recall that the dimensions multiply when you do the tensor product, so $L_1 \otimes L_2$ is still a line bundle.

Let's say that $c_1(L) = 1-L$ (there are other choices of Thom class), then we'd expect $c_1(L_1 \otimes L_2) = 1 - L_1 \otimes L_2$.

So, how do we express $c_1(L_1 \otimes L_2)$ in terms of a formal group law $F(x,y)$ where $x = c_1(L_1)$, and $y = c_1(L_2))$?

\begin{align*} F(x, y) & = x + y - xy \\ F(c_1(L_1), c_1(L_2)) & = c_1(L_1) + c_1(L_2) - c_1(L_1)c_1(L_2) \\ & = (1-L_1) + (1-L_2) - (1-L_1)(1-L_2) \\ & = 2 - L_1 - L_2 - 1 + L_1 + L_2 - L_1 \otimes L_2 \\ & = 1 - L_1 \otimes L_2 \end{align*}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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