Algebraic Topology – Computation of the Linking Invariant on Lens Spaces

at.algebraic-topologycohomologyhomology

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:
$$
lk(\Sigma)=\int _{L_n(p)}A\cup \beta(A)^n \in \mathbb{R}/\mathbb{Z}
$$

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

Best Answer

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.

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[Math] the hopf invariant of the hopf construction

This is my attempt at an answer. I think I might be off in justifying the second diagram. It's based on the proof in Bredon, but I think it might be a little simpler. Corrections are welcome, of course!

For convenience, we introduce some notation and conventions. We define subsets of $$S^n= \{ (c,t):c\in S^{n-1},t\in I \} /\sim$$ by $X=\{(c,t):t\leq 1/2\}$, $Y=\{(c,t):t\geq 1/2\}$, and $Z=\{(c,1/2)\}$. We define subsets of $$S^{2n-1}=\{ (a,t,b):a,b\in S^{n-1},t\in I\}$$ by $A=\{(a,t,b)\in S^{2n-1}:t\leq 1/2\}$ and $B=\{(a,t,b)\in S^{2n-1}:t\geq 1/2\}$. We define subsets of $A\cap B=\{(a,1/2,b)\}$ by $S^{n-1}_A = \{(a,1/2,b_0)\}$ and $S^{n-1}_B = \{(a_0,1/2,b)\}$ for fixed $a_0,b_0\in S^{n-1}$. We consider $S^{2n-1}$ as the boundary of $e^{2n}$; in this cell, $S^{n-1}_A$ is the boundary of a cell $e^n_A$ and $S^{n-1}_B$ is the boundary of a cell $e^n_B$. We write $K=S^n\cup_{h(g)}e^{2n}$ for the complex in which we must compute $H(h(g))$. Throughout, all cohomology is integral.

To compute cup products in $K$, we use the commutative diagram

H^n(K) x H^n(K) ---------> H^{2n}(K)
       ^             cup       ^
       |                       |
       |             cup       |
H^n(K,X) x H^n(K,Y) -----> H^{2n}(K,S^n)

which follows from the map $(K,\emptyset,\emptyset)\rightarrow (K,X,Y)$ and naturality; the vertical maps are isomorphisms by the long exact sequence in cohomology for pairs. Let the generator $x$ of $H^n(K)$ correspond to the generator $x_Y$ of $H^n(K,X)$ and the generator $x_X$ of $H^n(K,Y)$, so that $x_Y\cup x_X$ in $H^{2n}(K,S^n)$ corresponds to $x^2$ in $H^{2n}(K)$.

To understand the image of the cup product, we use the evident map $j:(e^{2n},A,B)\rightarrow (K,X,Y)$. We have the diagram

H^n(K,X)  =  H^n(S^n,X)  =  H^n(Y,Z)  =  H^{n-1}(Z)
   |                                       |
   | j^*                                   | j^*
   |                                       |
   V                                       V
H^n(e^{2n},A) = H^n(e^n_A,S^{n-1}_A) = H^{n-1}(S^{n-1}_A)

which commutes because of the naturality of the isomorphisms involved (namely, along the top: cellular cohomology, excision and homotopy invariance, lexseq for pairs; along the bottom: homotopy invariance, lexseq for pairs). By definition of degree, the map $j^\ast:H^{n-1}(Z)\rightarrow H^{n-1}(S^{n-1}_A)$ is multiplication by $\alpha$. Hence, so is the map $j^\ast : H^n(K,X)\rightarrow H^n(e^{2n},A)$. Similarly, the map $j^\ast:H^n(K,Y)\rightarrow H^n(e^{2n},B)$ is multiplication by $\beta$. Now, $j^\ast:H^{2n}(K,S^n)\rightarrow H^{2n}(e^{2n},S^{2n-1})$ is an isomorphism, and $j^\ast(x_Y\cup x_X)$ is $\alpha\beta$ times a generator of $H^{2n}(e^{2n},S^{2n-1})$. Thus $x_Y\cup x_X$ is $\alpha\beta$ times a generator of $H^{2n}(K,S^n)$. So by the first diagram, $H(h(g))=\alpha\beta$.

[Math] Intutive interpretation about Linking forms

This may be a good place to explain the well-known principle $$\text{intersection in the interior = linking in the boundary}$$ in an oriented $m$-dimensional manifold with boundary $(M,\partial M)$. Let $$f~:~(K,\partial K)\subset (M,\partial M)~,~g~:~(L,\partial L) \subset (M,\partial M)$$ be embeddings of oriented manifolds with boundary, such that $${\rm dim}(K)~=~k~,~{\rm dim}(L)~=~\ell~,~k+\ell~=~m~,~ f(\partial K) \cap g(\partial L)~=~\emptyset \subset \partial M~.$$ Assume there exists an isotopy (= homotopy through embeddings) rel $\partial K$ $$f_t~:~K \to M~~(0 \leqslant t \leqslant 1)$$ such that $f_0=f$ and $f_1(K)\subset \partial M$, with each $f_t(K), g(L)$ intersecting transversely in $M$, so that $f_t(K) \cap g(L)\subset M$ is a finite set with an intersection index $I(x)\in \{\pm 1\}$ at each point $x \in f_t(K) \cap g(L)$ according to the orientations. A continuity argument shows that the function $$\lambda~:~[0,1] \to {\mathbb Z}~:~t\mapsto \lambda(t)= \sum\limits_{x \in f_t(K) \cap g(L)}I(x)$$ is constant, so that $${\rm intersection}(f(K),g(L)\subset M)~=~\lambda(0)~=~\lambda(1)$$ $$=~{\rm linking}(f(\partial K),g(\partial L) \subset \partial M) \in {\mathbb Z}~.$$ This is best seen by drawing pictures for $(M,\partial M)=(D^2,S^1)$.

The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.

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