Equivariant complex $K$-theory of a real representation sphere

Question

Consider the one-point compactification of a $U(n)$-representation $V$, denoted by $S^V$. I want to caclulate $\tilde{K}_\ast^{U(n)}(S^V)$. When $V$ is a complex $U(n)$-representation, we can use the Equivariant Bott periodicity to infer that $\tilde{K}^G_0(S^V)\cong \tilde{K}^G_0(S^0)$. I want to know what happens when $V$ is a real $U(n)$-representation, more specifically, I want to identify the object $\tilde{K}^G_{\ast}(S^{\text{Ad}(U(n))})$, where $\text{Ad}(U(n))$ is the adjoint representation of $U(n)$.

My guess is, if the adjoint action of $U(n)$ is orientable, the Thom isomorphism gives the isomorphism to $K^{U(n)}_\ast(S^{|Ad(U(n))|})$ (?) on the real vector space $\text{Ad}(U(n))$ or the answer will be some twisted from of $\pi_\ast(K)[t,t^{-1}]$, where $t$ is a degree two element.

Answer 1

Here is one possible approach, which is specific to the adjoint representation. Let $X$ denote $U(n)$, regarded as a $U(n)$-space by the rule $g.x=gxg^{-1}$. Put $X_k=\{x\in X:\text{rank}(x-1)\leq k\}$, so we have a $U(n)$-equivariant filtration $$ \emptyset = X_{-1}\subset 1=X_0 \subset X_1 \subset \dotsb \subset X_n = X $$ Haynes Miller proved that there is a stable splitting $\Sigma^\infty X_+\simeq\bigvee_{i=0}^nX_i/X_{i-1}$, and one can check that this works equivariantly. For $u\in\mathfrak{u}(n)$ we have the Cayley transform $g=(u+1)(u-1)^{-1}$ which lies in $X_n\setminus X_{n-1}$. This construction gives a homeomorphism $\mathfrak{u}(n)\simeq X_n\setminus X_{n-1}$, and we can pass to the one-point compactification to get a homeomorphism $S^{\mathfrak{u}(n)}\simeq X_n/X_{n-1}$. More generally, suppose we have a subspace $V\leq\mathbb{C}^n$ of dimension $k$, and an antihermitian endomorphism $u$ of $V$; this gives an element $g=((u+1)(u-1)^{-1})\oplus 1_{V^\perp}$ lying in $X_k\setminus X_{k-1}$. If we let $G_k=U(n)/(U(k)\times U(n-k))$ be the Grassmannian of $k$-planes in $\mathbb{C}^n$, and $T$ the tautological bundle over $G_k$, and $\mathfrak{u}(T)$ the associated bundle of antihermitian endomorphisms, and $G_k^{\mathfrak{u}(T)}$ the corresponding Thom space, we find that $X_k/X_{k-1}\simeq G_k^{\mathfrak{u}(T)}$. Thus, Miller's splitting takes the form $$ \Sigma^\infty X_+ \simeq \bigvee_{k=0}^n G_k^{\mathfrak{u}(T)}. $$ Thus, the group $K_{U(n)}^*(X)$ splits as a direct sum of pieces, of which the top piece is the group $K_{U(n)}^*(S^{\mathfrak{u}(n)})$ that you asked about, and the other pieces are closely related.

I think that there is literature about $K_{U(n)}^*(X)$ but I do not remember at the moment what it says. One point is that we can let $I$ be the augmentation ideal in $K_{U(n)}^*(\text{point})=R(U(n))[v,v^{-1}]$ and form the completion $K_{U(n)}^*(X)^\wedge_I$. The Atiyah-Segal completion theorem identifies this with $K^*(EU(n)\times_{U(n)}X)$. If I remember correctly we can also identify $EU(n)\times_{U(n)}X$ with the free loop space $LBU(n)=\text{Map}(S^1,BU(n))$.