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.

  • $\begingroup$ 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


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.


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