3
$\begingroup$

Reference: H. Blaine Lawson, Spin Geometry, Page 72

Problem: Here Remark 10.5 states an internal symmetry in the KR-theory that for any compact space $X$ there are isomorphisms $$KR(X\times S^{0,p}) \cong KR^{-2p}(X\times S^{0,p}) ~~~~~~~~~~~~~(1)$$ for $p=1,2,4$. If $\mathbb R^{r,s} =\mathbb R^r \oplus \mathbb R^s$ be the Real linear space with involution $c(x,y)=(x,-y)$, then we denote $S^{r,s}\equiv \{(x,y)\in\mathbb R^{r,s}: ||x||^2+||y||^2 = 1\}$

The book claims that the case $p=1$ gives the (general) complex Bott Periodicity Theorem which states that $\mu_\xi:K^{-i}(X) \to K^{-i-2}(X) $ given by module multiplication by $\xi$ which is the generator of the ring $K^{-*}(pt)\cong \mathbb Z[\xi]$, the isomorphism which is stated by the (special) Bott Periodicity Theorem; and the book also claims that the case $p=4$ gives the real periodicity theorem.

For $p=1$ the isomorphism (1) gives $KR(X\times S^{0,1}) \cong KR^{-2}(X\times S^{0,1})$. Note that by the above definition $S^{0,1}$ is a set $\{p,q\}$of two points with the involution map $c_0: p\mapsto q,~q\mapsto p$. So I want to consider the involution map $c:X\times S^{0,1} \to X\times S^{0,1}$ given by $c=Id_X \times c_0$. However I cannot see any relation to the complex Bott Periodicity Theorem as above.

Moreover, in general, how to recover the K-theory from the KR-theory? In this book Remark 10.2 explains how to recover the KO-theory only.

$\endgroup$
1
  • $\begingroup$ A look into Atiyah's ''K-Theory and Reality'' should answer most questions about these issues. $\endgroup$ Jan 23, 2017 at 22:25

1 Answer 1

7
$\begingroup$

Unless I am mistaken, $KU(X) = KR(X \times S^{0,1})$ for $X$ an ordinary space. To see why this is reasonable, unpack $X\times S^{0,1} = X\sqcup X$ with the obvious involution and check that Real vector bundles on $X \sqcup X$ are the same as ordinary (complex) vector bundles on (the first copy of) $X$. Indeed, given a vector bundle on the first $X$, put its conjugate on the second copy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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