5
$\begingroup$

Let $L_n(p)$ be the $2n+1$ dimensional Lens space $$ S^{2n+1}/\mathbb{Z}_p $$ where the action is given as $z_i\rightarrow e^{\frac{2\pi}{p}}z_i$, $i=1,...,n+1$, with $z_i$ the coordinates of $\mathbb{C}^{n+1}$ such that $S^{2n+1}$ is $|z_1|^2+...+|z_{n+1}|^2=1$. For $k\neq 0,2n+1$ the homology groups with coefficients in a commutative ring $R$ are $$ H_k(L_n(p),R)=\left\{\begin{array}{cc} R/pR & \text{if $k$ is odd}\\ T_p(R) & \text{if $k$ is even } \end{array}\right. $$ where $T_p(R)\subset R$ is the $p-$torsion subgroup: $$ T_p(R)=\left\{x\in R \ | \ px=0\right\} \ . $$ Since $R$ is an abelian group, we can consider $R^{\vee}$, its Pontryagin. Fix an extension $\Gamma$ of $R$ by $R^{\vee}$: $$ 1\rightarrow R^{\vee}\rightarrow \Gamma \rightarrow R\rightarrow 1 \ . $$ This induces a long exact sequence of homology and cohomology groups, with connecting homomorphisms given by the Bockstein maps. For instance in cohomology $\beta :H^k(L_n(p),R)\rightarrow H^{k+1}(L_n(p),R^{\vee})$.

Given $\Sigma \in H_{2n}(L_n(p),R)$ we consider its Poincare' dual cocycle $A=PD(\Sigma)\in H^1(L_n(p),R)$, and its image under Bockstein $\beta(A)\in H^2(L_n(p),R^{\vee})$. Since $R^{\vee}$ is also a ring, the cup product associated with the product $R^{\vee}\times R^{\vee}\rightarrow R^{\vee}$ allows to construct a class $\beta(A)^n\in H^{2n}(L_n(p),R^{\vee})$, and the natural pairing $R\times R^{\vee}\rightarrow \mathbb{R}/\mathbb{Z}$ allows the construction of the self-linking invariant: $$ lk(\Sigma)=\int _{L_n(p)}A\cup \beta(A)^n \in \mathbb{R}/\mathbb{Z} $$

  1. Preliminary question: how this self-linking invariant is related with the more standard linking form which only uses the sequence $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$?

  2. Let us first fix $n=1$, $p=2$ so that $L_1(p)=\mathbb{RP}^3$, and $R=\mathbb{Z}_2$. Since $T_2(\mathbb{Z}_2)=\mathbb{Z}_2$ there is precisely one 2-cycle $\Sigma \in H_2(L_1(2),\mathbb{Z}_2)$, and one 1-cycle $\gamma\in H_1(L_1(2),\mathbb{Z}_2)$. How can I compute $$ lk(\Sigma)=\int _{L_1(2)}A\cup \beta(A) $$ explicitely? Moreover I have the intuition that $\beta(A)$ should be the Poincare' dual of $\gamma$, but I don't know how to prove this.

  3. For general $n$ and $p$ fix $R=\mathbb{Z}_p$, so that $R/pR=T_p(R)=\mathbb{Z}_p$, and again choose one generator $\Sigma $ of $H_{2n}(L_n(p),\mathbb{Z}_p)$. How can I compute $lk(\Sigma)$? By extending the intuition before I also guess that, fixing a generator $\gamma$ of $H_1(L_n(p),\mathbb{Z}_p)$, $\beta(A)$ is the Poincare' dual of $\beta(A)^n$. Is this correct?

$\endgroup$
2
  • 1
    $\begingroup$ Why is the extension $\Gamma$ not appearing in your guesses? Do you have a particular one in mind? $\endgroup$ May 14 at 15:10
  • $\begingroup$ That's a good point. Indeed I think the guesses should be refined somehow, but I don't know how. In some way the extension appears in the definition of the Bockstein map, but it is certainly true that the Poincarè dual cannot depend on it. If the extension is trivial (split sequence) so is the Bockstein and the guess is certainly wrong. I am assuming a non-trivial extension, and maybe there is some factor between the Poincarè dual and $\beta(A)$.... $\endgroup$ May 14 at 15:22

1 Answer 1

5
$\begingroup$

In my thesis, I gave the calculation of the linking form on homology, which is equivalent to the question you asked. I credited the calculation to de Rham (Sur L'analysis situs des varietés a n dimensions, J. Math. Pures et Appl., 10 (1931), 115-200.) See Proposition 4 in Imbedding punctured lens spaces and connected sums. Pacific J. Math. 113 (1984), no. 2, 481–491.

The answer is given for a lens space $L(m;q_1,\ldots,q_{2k})$ of dimension $4k-1$; I imagine that it's similar for dimension $4k+1$. With respect to a certain generator $e$ of $H_{2k-1}(L)$, de Rham showed $\lambda(e,e) = \frac1m q_1\cdots q_k \cdot r_{k+1}\cdots r_{2k}$ where $q_jr_j \equiv 1 \pmod{m}$. You'd have to look in de Rham's paper to find the details.

$\endgroup$
1
  • $\begingroup$ Thank you for the answer. Could you maybe explain in which sense this is equivalent to computing $A\cup \beta(A)^n$ on the $X$? In particular I am confused by the fact that in the definition of the linking form there is no reference to the groups of coefficients, while the Bockstein depends on the short exact sequence... Do you also have some comment on the Poincarè dual? $\endgroup$ May 16 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.