Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am interested in the real and imaginary part of the complex polylogarithm $$L_{k+1}(\zeta):=Re(\frac 1 {i^k}\sum_{m=1}^\infty \frac{\zeta^m}{m^{k+1}}),$$ where $\zeta$ is a primitive $n$-th root of unity and $n$ not necessarily prime. Now we can form the $n\times n$-matrix $$A_{i,l}:=(L_{k+1}(\zeta^{il})).$$ My question is: What is the rank of this matrix over $\mathbb{Q}$?

Because of the topological background this question arose in I expect this rank to be the rank of the $K$-group $K_{2k+1}(\mathbb{Z}[\mathbb{Z}/n]),$ which is $\frac{n-1} 2$ if $k$ and $n$ are odd, $\frac{n+1} 2$ if $k$ is even and $n$ is odd and $\frac{n-2} 2$ and $\frac{n+2} 2$ if $n$ is even and $k$ is odd and even respectively.

The background is the following: We will look at the classifying space of $h$-cobordism bundles of high-dimensional lens spaces $\mathcal{H}(B\mathbb{Z}/n).$ I can explicitly construct elements $\Delta^i\in \pi_{2k}\mathcal{H}(B\mathbb{Z}/n)$ for $i=1,\ldots, n$ and I want to show that these rationally span this homotopy group. We know $$\mathcal{H}(B\mathbb{Z}/n)\simeq_\mathbb{Q} \Omega A(B\mathbb{Z}/n)\simeq_\mathbb{Q}\Omega K(\mathbb{Z}[\mathbb{Z}/n]).$$ Furthermore we have the torsion vector as a group homomorphism $$T:\pi_{2k}\mathcal{H}(B\mathbb{Z}/n)\otimes \mathbb{Q} \to H^{2k}(S^{2k};\mathbb{R}^n)\cong \mathbb{R}^n; \Delta \mapsto (\tau^{IK}_k(\Delta,\zeta^i)),$$ where $=1,\ldots, n$ and $\tau^{IK}_k(\Delta,\zeta)$ is the twisted higher Reidemeister torsion or Igusa-Klein torsion in degree $2k$ of $\Delta$ with twisted coefficients $\zeta.$ A calculation now shows $$\tau_k^{IK}(\Delta^i,\zeta)=L_{k+1}(\zeta^i)$$ and thereby the rank of the matrix $A$ corresponds to the rank of the span of the elements $\Delta^i$ in $\pi_{2k}(\mathcal{H}(B\mathbb{Z}/n).$

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.