# Cyclic cohomology of a field $k$

Let $$A$$ be a $$k$$-algebra where $$k$$ is a field. Define $$C^n(A):=\text{Hom}_k(A^{\otimes n+1}, k)$$, where $$A^{\otimes n+1}$$ is the $$n$$-fold $$k$$ tensor product of $$A$$ with itself. Then the cyclic cohomology $$HC^{\ast}(A)$$ of $$A$$ is the cohomology of the total complex of $$\require{AMScd}$$ $$\begin{CD} \cdots @. \cdots @. \cdots \\ @AAA @AAA @AAA \\ C^2(A) @>B>> C^1(A) @>B>> C^0(A) \\ @AbAA @AbAA \\ C^1(A) @>B>> C^0(A) \\ @AbAA \\ C^0(A) \end{CD}$$ where $$b:C^n(A)\to C^{n+1}(A)$$ is the Hochschild coboundary map and $$B:C^{n+1}(A)\to C^{n}(A)$$ is given by \begin{align} B(f)(a_0\otimes\dots\otimes a_n) &= \sum_{i=0}^n(-1)^{ni}f(1\otimes a_i\otimes\dots\otimes a_n\otimes a_0\otimes\dots\otimes a_{i-1}) \\ &\quad -(-1)^{ni}f(a_i\otimes 1\otimes a_{i+1}\otimes\dots\otimes a_n\otimes a_0\otimes\dots\otimes a_{i-1}) \end{align} I'm trying to figure out what $$HC^{\ast}(k)$$ is. According to Loday in Cyclic Homology (pg. 74) it is "immediate" that $$HC^{2n}(k) = k$$ and $$HC^{2n+1}(k) = 0$$ for $$n\ge 0$$. Unfortunately though, this isn't immediate to me and he doesn't give any other details. I know that when $$A = k$$ then the complex above becomes $$\begin{CD} \cdots @. \cdots @. \cdots \\ @A0AA @AidAA @A0AA \\ C^2(k) @>0>> C^1(k) @>0>> C^0(k) \\ @AidAA @A0AA \\ C^1(k) @>0>> C^0(k) \\ @A0AA \\ C^0(k) \end{CD}$$ since $$C^n(k)\simeq C^0(k)$$ for all $$n\ge 0$$. So the Hochschild coboundary $$b:C^n(k)\to C^{n+1}(k)$$ is identity when $$n$$ is odd and the zero map when $$n$$ is even. The map $$B$$ is always the zero map.

I understand why the cyclic cohomology is $$2$$-periodic, and I understand why the odd cyclic cohomology is $$0$$. What I don't understand is why $$HC^{2n}(k) = k$$. Reading straight off the complex you get $$HC^0(k) = \text{ker}(0) = C^0(k) = \text{Hom}_k(k, k)$$, but I don't understand how to identify $$\text{Hom}_k(k, k)$$ with $$k$$. From Loday's use of the word "immediate" I'm guessing there is some quick trick that makes identifying these two a simple task? If there is one I don't know what it is. Any help would be appreciated!

Edit: I've had a thought about how to show that $$\text{Hom}_k(k, k)\simeq k$$. So since $$\text{Hom}_k(k, k)$$ has the structure of a $$k$$-bimodule and $$k$$ is a $$k$$-module, then for each $$f\in \text{Hom}_k(k, k)$$ and $$a\in k$$ we have

$$f(a) = f(a1) = af(1)$$

so that each $$f$$ is uniquely determined by how it acts on $$1$$. Then for any $$a^\prime\in k$$ define the map $$f_{a^\prime}(a) = aa^\prime$$. This is a well defined $$k$$-module homomorphism. Then we have

$$f_{a^\prime}(a) = af_{a^\prime}(1)$$

and since $$k$$ is a field then every $$a\in k$$ has a multiplicative inverse $$a^{-1}$$ and so

$$f_{a^\prime}(1) = a^{-1}af_{a^\prime}(1) = a^{-1}aa^\prime =a^\prime$$

So the mapping that sends $$f\mapsto f(1)$$ is a module isomorphism. Is this a valid argument?

Edit 2: Think I may have found a more elegant solution. So, Theorem 2.4 in these notes states that if $$M$$ is a free $$k$$-module of finite rank $$n$$ then the dual space $$M^{\lor}:=\text{Hom}_k(M, k)$$ is also a free $$k$$-module of rank $$n$$. Therefore $$k^{\lor}=\text{Hom}_k(k, k)$$ is a rank $$1$$ free $$k$$-module, which is isomorphic to $$k$$.

Theorem 2.4 in these notes states that if $$M$$ is a free $$k$$-module of finite rank $$n$$ then the dual space $$M^{\lor}:=\text{Hom}_k(M, k)$$ is also a free $$k$$-module of rank $$n$$. Therefore $$k^{\lor}=\text{Hom}_k(k, k)$$ is a rank $$1$$ free $$k$$-module, which is isomorphic to $$k$$.
Therefore $$HC^{2n}(k) = \text{Hom}_k(k, k) \simeq k$$ as required.
• $\text{Hom}_k(k, k) \cong k$ because a $k$-linear map $k \to k$ is uniquely and freely determined by what it does to the basis vector $1 \in k$. Oct 8 '20 at 0:56