I am trying to understand the assembly map

$$\pi_i ((BC_2)_+ \wedge A( \ast )) \rightarrow A_i( BC_2 ) $$

in low degrees for the space $BC_2 = \mathbb R P^\infty$ in Waldhausen $A$-theory. I know we have a splitting $A( X ) \simeq \Sigma^\infty X_+ \vee \text{Wh}^{\text{DIFF}}(X)$ and in the case of the point, $\text{Wh}^{\text{DIFF}}(\ast)$ is a 3-connective spectrum, with $\text{Wh}_3^{\text{DIFF}}(\ast) = \mathbb Z/2$ and $\text{Wh}_4^{\text{DIFF}}(\ast) = 0$. This means, the question of the injectivity of the assembly reduces to the assembly in $\text{Wh}^{\text{DIFF}}$, which makes it easy to show that the assembly is injective up till degree 3.

In degree 4 we're left with a term $ \pi_4 ((BC_2)_+ \wedge \text{Wh}^{\text{DIFF}}(\ast)) = H_1( C_2; \mathbb Z/2) = \mathbb Z/2$. The image of this term seems to be of purely topological nature, as:

  1. It, more or less by definition, doesn't come from the stable homotopy group $\pi_4^{st} (BC_2)$
  2. It is killed under the morphism $A( BC_2 ) \rightarrow A( \ast )$ which is induced by the map that sends the generator $t \in C_2$ to $-1$.
  3. It is killed under passage to $K$-theory of any discrete ring, i.e. the maps $A( BC_2) \rightarrow K ( R [C_2])$ can't detect it.

Which brings me to my question: Has $A_4( BC_2 )$ already been computed and what can we say about the image of the assembly?


1 Answer 1


Not an answer, but a possible approach: Using Dundas' cartesian square with corners $A(*)$, $K(Z)$, $TC(*)$ and $TC(Z)$ you can see that $Wh^{Diff}_3(*) = Z/2$ comes from $TC_4(Z) = Z/8$ (plus odd torsion) via the Mayer-Vietoris connecting homomorphism $\partial$. Hence $H_1(BC_2; Wh^{Diff}_3(*)) = Z/2$ also comes from $H_1(BC_2; TC_4(Z)) = Z/2$ via the homomorphism induced by $\partial$.

I think there is a commutative square with corners $H_1(BC_2; TC_4(Z))$, $H_1(BC_2; A_3(*))$, $TC_5(Z[C_2])$ and $A_4(BC_2)$. Hence you might try to calculate the assembly map $H_1(BC_2; TC_4(Z)) \to TC_5(Z[C_2])$. If its image in $TC_5(Z[C_2])$ is nonzero, and not in the combined image from $K_5(Z[C_2])$ and $TC_5(BC_2)$, then the assembly map to $A_4(BC_2)$ will be injective.

I think calculating the 2-torsion in $TC_5(Z[C_2])$ (or $K_5(Z_2[C_2])$) is the harder part here. Understanding $TC_5(BC_2)$ should be easy from the Boekstedt-Hsiang-Madsen formula. Identifying the image from $K_5(Z[C_2])$ might also be hard.

  • $\begingroup$ I have in fact just been brooding over your paper "Two-primary algebraic K-theory of pointed spaces" for a solution! I will not check this answer as accepted for now, try those calculations, and accept it in case they turn out to be fruitful. $\endgroup$ Jan 8, 2022 at 12:40

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