Linking form

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Introduction to linking forms

After Poincaré and Lefschetz, a closed oriented manifold $N^{n}$${{Stub}} == Introduction to linking forms== <wikitex>; After Poincaré and Lefschetz, a closed oriented manifold $N^{n}$ has a bilinear intersection form defined on its homology: $$I_N \colon H_k(N;\mathbb{Z}) \times H_{n-k}(N;\mathbb{Z}) \to \mathbb{Z}; ([p],[q]) \mapsto \langle p, q \rangle$$ such that $$I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$$ Given a ${k}$--chain $p \in C_{k}(N;\mathbb{Z})$ and an $(n-k)$--chain $q \in C_{n-k}(N;\mathbb{Z})$ which is transverse to $q$, the signed count of the intersections between $p$ and $q$ gives an intersection number $\langle\, p \, , \, q\, \rangle \in \mathbb{Z}$. By bilinearity, the intersection form vanishes on the torsion part of the homology. The analogue of the intersection pairing for the torsion part of the homology of a closed oriented manifold $N^n$ is the bilinear $\mathbb{Q}/\mathbb{Z}$--valued linking form, which is due to Seifert: $$L_N \colon TH_{\ell}(N;\mathbb{Z}) \times TH_{n-\ell-1}(N;\mathbb{Z}) \to \mathbb{Q}/\mathbb{Z}$$ such that $$L_N(x,y) = (-)^{\ell(n-\ell-1)}L_N(y,x).$$ Given $[x] \in TH_\ell(N;\mathbb{Z})$ and $[y] \in TH_{n-\ell-1}(N;\mathbb{Z})$ represented by cycles $x \in C_\ell(N;\mathbb{Z})$ and $y \in C_{n-\ell-1}(N,\mathbb{Z})$, let $w \in C_{\ell+1}(N;\mathbb{Z})$ be such that $\partial w = sy$, for some $s \in \mathbb{Z}$. Then we define: $$L_N([x],[y]) := \langle x, w \rangle/s \in \mathbb{Q}/\mathbb{Z}.$$ The resulting element is independent of the choices of $x,y,w$ and $s$. As an example, let $N = \mathbb{RP}^3$, so that $\ell=1$ and $n=3$. Now $H_1(\mathbb{RP}^3;\mathbb{Z}) \cong \mathbb{Z}_2$. Let $\theta \in H_1(\mathbb{RP}^3;\mathbb{Z})$ be the non--trivial element. To compute the linking $L_{\mathbb{RP}^3}(\theta,\theta)$, consider $\mathbb{RP}^3$ modelled as $D^3/\sim$, with antipodal points on $\partial D^2$ identified, and choose two representative N^{n}$ has a bilinear intersection form defined on its homology:

$\displaystyle I_N \colon H_k(N;\mathbb{Z}) \times H_{n-k}(N;\mathbb{Z}) \to \mathbb{Z}; ([p],[q]) \mapsto \langle p, q \rangle$ $$\displaystyle I_N \colon H_k(N;\mathbb{Z}) \times H_{n-k}(N;\mathbb{Z}) \to \mathbb{Z}; ([p],[q]) \mapsto \langle p, q \rangle$$

such that

$\displaystyle I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$ $$\displaystyle I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$$

Given a ${k}$${k}$--chain $p \in C_{k}(N;\mathbb{Z})$$p \in C_{k}(N;\mathbb{Z})$ and an $(n-k)$$(n-k)$--chain $q \in C_{n-k}(N;\mathbb{Z})$$q \in C_{n-k}(N;\mathbb{Z})$ which is transverse to $q$$q$, the signed count of the intersections between $p$$p$ and $q$$q$ gives an intersection number $\langle\, p \, , \, q\, \rangle \in \mathbb{Z}$$\langle\, p \, , \, q\, \rangle \in \mathbb{Z}$.

By bilinearity, the intersection form vanishes on the torsion part of the homology. The analogue of the intersection pairing for the torsion part of the homology of a closed oriented manifold $N^n$$N^n$ is the bilinear $\mathbb{Q}/\mathbb{Z}$$\mathbb{Q}/\mathbb{Z}$--valued linking form, which is due to Seifert:

$\displaystyle L_N \colon TH_{\ell}(N;\mathbb{Z}) \times TH_{n-\ell-1}(N;\mathbb{Z}) \to \mathbb{Q}/\mathbb{Z}$ $$\displaystyle L_N \colon TH_{\ell}(N;\mathbb{Z}) \times TH_{n-\ell-1}(N;\mathbb{Z}) \to \mathbb{Q}/\mathbb{Z}$$

such that

$\displaystyle L_N(x,y) = (-)^{\ell(n-\ell-1)}L_N(y,x).$ $$\displaystyle L_N(x,y) = (-)^{\ell(n-\ell-1)}L_N(y,x).$$

Given $[x] \in TH_\ell(N;\mathbb{Z})$$[x] \in TH_\ell(N;\mathbb{Z})$ and $[y] \in TH_{n-\ell-1}(N;\mathbb{Z})$$[y] \in TH_{n-\ell-1}(N;\mathbb{Z})$ represented by cycles $x \in C_\ell(N;\mathbb{Z})$$x \in C_\ell(N;\mathbb{Z})$ and $y \in C_{n-\ell-1}(N,\mathbb{Z})$$y \in C_{n-\ell-1}(N,\mathbb{Z})$, let $w \in C_{\ell+1}(N;\mathbb{Z})$$w \in C_{\ell+1}(N;\mathbb{Z})$ be such that $\partial w = sy$$\partial w = sy$, for some $s \in \mathbb{Z}$$s \in \mathbb{Z}$. Then we define:

$\displaystyle L_N([x],[y]) := \langle x, w \rangle/s \in \mathbb{Q}/\mathbb{Z}.$ $$\displaystyle L_N([x],[y]) := \langle x, w \rangle/s \in \mathbb{Q}/\mathbb{Z}.$$

The resulting element is independent of the choices of $x,y,w$$x,y,w$ and $s$$s$.

As an example, let $N = \mathbb{RP}^3$$N = \mathbb{RP}^3$, so that $\ell=1$$\ell=1$ and $n=3$$n=3$. Now $H_1(\mathbb{RP}^3;\mathbb{Z}) \cong \mathbb{Z}_2$$H_1(\mathbb{RP}^3;\mathbb{Z}) \cong \mathbb{Z}_2$. Let $\theta \in H_1(\mathbb{RP}^3;\mathbb{Z})$$\theta \in H_1(\mathbb{RP}^3;\mathbb{Z})$ be the non--trivial element. To compute the linking $L_{\mathbb{RP}^3}(\theta,\theta)$$L_{\mathbb{RP}^3}(\theta,\theta)$, consider $\mathbb{RP}^3$$\mathbb{RP}^3$ modelled as $D^3/\sim$$D^3/\sim$, with antipodal points on $\partial D^2$$\partial D^2$ identified, and choose two representative $1$$1$--chains $x$$x$ and $y$$y$ for $\theta$$\theta$. Let $x$$x$ be the straight line between north and south poles and let $y$$y$ be half of the equator. Now $2y = \partial w$$2y = \partial w$, where $w \in C_2(\mathbb{RP}^3;\mathbb{Z})$$w \in C_2(\mathbb{RP}^3;\mathbb{Z})$ is the 2-disk whose boundary is the equator. We see that $\langle x,w \rangle = 1$$\langle x,w \rangle = 1$, so that

$\displaystyle L_{\mathbb{RP}^3}(\theta,\theta) = L_{\mathbb{RP}^3}([x],[y]) = \langle x,y \rangle/2 = 1/2.$ $$\displaystyle L_{\mathbb{RP}^3}(\theta,\theta) = L_{\mathbb{RP}^3}([x],[y]) = \langle x,y \rangle/2 = 1/2.$$

References

$--chains $x$ and $y$ for $\theta$. Let $x$ be the straight line between north and south poles and let $y$ be half of the equator. Now y = \partial w$, where $w \in C_2(\mathbb{RP}^3;\mathbb{Z})$ is the 2-disk whose boundary is the equator. We see that $\langle x,w \rangle = 1$, so that $$L_{\mathbb{RP}^3}(\theta,\theta) = L_{\mathbb{RP}^3}([x],[y]) = \langle x,y \rangle/2 = 1/2.$$ == References == {{#RefList:}} [[Category:Theory]] [[Category:Definitions]]N^{n} has a bilinear intersection form defined on its homology:

$\displaystyle I_N \colon H_k(N;\mathbb{Z}) \times H_{n-k}(N;\mathbb{Z}) \to \mathbb{Z}; ([p],[q]) \mapsto \langle p, q \rangle$ $$\displaystyle I_N \colon H_k(N;\mathbb{Z}) \times H_{n-k}(N;\mathbb{Z}) \to \mathbb{Z}; ([p],[q]) \mapsto \langle p, q \rangle$$

such that

$\displaystyle I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$ $$\displaystyle I_N(x,y) = (-)^{k(n-k)}I_N(y,x).$$

Given a ${k}$${k}$--chain $p \in C_{k}(N;\mathbb{Z})$$p \in C_{k}(N;\mathbb{Z})$ and an $(n-k)$$(n-k)$--chain $q \in C_{n-k}(N;\mathbb{Z})$$q \in C_{n-k}(N;\mathbb{Z})$ which is transverse to $q$$q$, the signed count of the intersections between $p$$p$ and $q$$q$ gives an intersection number $\langle\, p \, , \, q\, \rangle \in \mathbb{Z}$$\langle\, p \, , \, q\, \rangle \in \mathbb{Z}$.

By bilinearity, the intersection form vanishes on the torsion part of the homology. The analogue of the intersection pairing for the torsion part of the homology of a closed oriented manifold $N^n$$N^n$ is the bilinear $\mathbb{Q}/\mathbb{Z}$$\mathbb{Q}/\mathbb{Z}$--valued linking form, which is due to Seifert:

$\displaystyle L_N \colon TH_{\ell}(N;\mathbb{Z}) \times TH_{n-\ell-1}(N;\mathbb{Z}) \to \mathbb{Q}/\mathbb{Z}$ $$\displaystyle L_N \colon TH_{\ell}(N;\mathbb{Z}) \times TH_{n-\ell-1}(N;\mathbb{Z}) \to \mathbb{Q}/\mathbb{Z}$$

such that

$\displaystyle L_N(x,y) = (-)^{\ell(n-\ell-1)}L_N(y,x).$ $$\displaystyle L_N(x,y) = (-)^{\ell(n-\ell-1)}L_N(y,x).$$

Given $[x] \in TH_\ell(N;\mathbb{Z})$$[x] \in TH_\ell(N;\mathbb{Z})$ and $[y] \in TH_{n-\ell-1}(N;\mathbb{Z})$$[y] \in TH_{n-\ell-1}(N;\mathbb{Z})$ represented by cycles $x \in C_\ell(N;\mathbb{Z})$$x \in C_\ell(N;\mathbb{Z})$ and $y \in C_{n-\ell-1}(N,\mathbb{Z})$$y \in C_{n-\ell-1}(N,\mathbb{Z})$, let $w \in C_{\ell+1}(N;\mathbb{Z})$$w \in C_{\ell+1}(N;\mathbb{Z})$ be such that $\partial w = sy$$\partial w = sy$, for some $s \in \mathbb{Z}$$s \in \mathbb{Z}$. Then we define:

$\displaystyle L_N([x],[y]) := \langle x, w \rangle/s \in \mathbb{Q}/\mathbb{Z}.$ $$\displaystyle L_N([x],[y]) := \langle x, w \rangle/s \in \mathbb{Q}/\mathbb{Z}.$$

The resulting element is independent of the choices of $x,y,w$$x,y,w$ and $s$$s$.

As an example, let $N = \mathbb{RP}^3$$N = \mathbb{RP}^3$, so that $\ell=1$$\ell=1$ and $n=3$$n=3$. Now $H_1(\mathbb{RP}^3;\mathbb{Z}) \cong \mathbb{Z}_2$$H_1(\mathbb{RP}^3;\mathbb{Z}) \cong \mathbb{Z}_2$. Let $\theta \in H_1(\mathbb{RP}^3;\mathbb{Z})$$\theta \in H_1(\mathbb{RP}^3;\mathbb{Z})$ be the non--trivial element. To compute the linking $L_{\mathbb{RP}^3}(\theta,\theta)$$L_{\mathbb{RP}^3}(\theta,\theta)$, consider $\mathbb{RP}^3$$\mathbb{RP}^3$ modelled as $D^3/\sim$$D^3/\sim$, with antipodal points on $\partial D^2$$\partial D^2$ identified, and choose two representative $1$$1$--chains $x$$x$ and $y$$y$ for $\theta$$\theta$. Let $x$$x$ be the straight line between north and south poles and let $y$$y$ be half of the equator. Now $2y = \partial w$$2y = \partial w$, where $w \in C_2(\mathbb{RP}^3;\mathbb{Z})$$w \in C_2(\mathbb{RP}^3;\mathbb{Z})$ is the 2-disk whose boundary is the equator. We see that $\langle x,w \rangle = 1$$\langle x,w \rangle = 1$, so that

$\displaystyle L_{\mathbb{RP}^3}(\theta,\theta) = L_{\mathbb{RP}^3}([x],[y]) = \langle x,y \rangle/2 = 1/2.$ $$\displaystyle L_{\mathbb{RP}^3}(\theta,\theta) = L_{\mathbb{RP}^3}([x],[y]) = \langle x,y \rangle/2 = 1/2.$$