Linking number definitions: equivalent or not?

Yes, these are equivalent. (In the following, $S^3$ and $\mathbb{R}^3$ are pretty much interchangeable. I was trying to give the whole story for linking numbers, and while it is incomplete, hopefully it sheds some light on the subject for you.)

The second linking number can be thought of in the following way: $H^1(S^3-C_2)\cong H_1(C_2)\cong\mathbb{Z}$ by Alexander duality, with the generator being the dual of the fundamental class $[C_2]\in H_1(C_2)$, and then by the universal coefficient theorem there is an isomorphism $$h:H^1(S^3-C_2)\to \operatorname{Hom}(H_1(S^3-C_2),\mathbb{Z}),$$ which can also be thought of as the evaluation pairing $H^1(S^3-C_2)\otimes H_1(S^3-C_2)\to \mathbb{Z}$. Then $\operatorname{link}(C_1,C_2)=h([C_2]^*)(i_*[C_1])$, where $i_*[C_1]$ is the homology class of the curve $C_1$ in $H_1(S^3-C_2)$. In terms of de Rham cohomology, $[C_2]^*$ corresponds to a $1$-form $\omega_2$, and $$h([C_2]^*)(i_*[C_1]) = \int_{C_1} \omega_2.$$ Or, put another way, if $[C_2]_*$ denotes the generator of $H_1(S^3-C_2)$ (chosen to be dual to $[C_2]^*$ with respect to the above pairing) then $\operatorname{link}(C_1,C_2)[C_2]_*=i_*[C_1]$.

By Poincare duality (and excision), $H^1(S^3-C_2)\cong H_2(S^3,C_2)$, and the dual of $[C_2]^*$ can be represented by an oriented surface $\Sigma_2\subset S^3$ with $\partial\Sigma_2=C_2$ with the correct induced orientation. In knot theory, this is called a Seifert surface for $C_2$. Then, $\operatorname{link}(C_1,C_2)=\Sigma_2\cdot C_1$, which is the algebraic intersection number between the surface and the curve with the result in $H_0(S^3-C_2)\cong\mathbb{Z}$. The way this works is you use a standard transversality argument to get a nearby $\Sigma_2$ that is transverse to $C_1$, so then $|\Sigma_2\cap C_1|$ is finite, and then you use the local orientation of $\Sigma_2$ concatenated with that of $C_1$ to determine the sign of the intersection, then add up all the signs to get the linking number.

Why is this version of the linking number symmetric? Given two Seifert surfaces, the algebraic intersection $\Sigma_1\cdot\Sigma_2$ gives an element of $H_1(S^3,C_1\cup C_2)$, represented by oriented arcs of intersection with endpoints on $C_1$ and $C_2$; any loops or arcs from a $C_i$ to itself are nullhomologous. (This is in terms of the Poincare dual; in terms of cohomology, we calculate the cup product $[C_1]^*\smile [C_2]^*\in H^2(S^3-(C_1\cup C_2))$, and in terms of de Rham cohomology the $2$-form $\omega_1\wedge\omega_2$.) One can check that $\partial(\Sigma_1\cdot\Sigma_2)$ in $H_1(C_1\cup C_2)$ is $\partial\Sigma_1\cdot\Sigma_2+\Sigma_1\cdot\partial\Sigma_2=\Sigma_2\cdot C_1+\Sigma_1\cdot C_2$. Arcs connect points of $\Sigma_1\cap C_2$ to $\Sigma_2\cap C_1$, and by using the fact $S_3$ is oriented we can see the arc gives that both intersections have the same sign, and so $\Sigma_1\cdot C_2$ and $\Sigma_2\cdot C_1$ are equal (up $H_0(S^3-C_1)\cong H_0(S^3-C_2)$).

One way to show the degree of the Gauss map equals this version of the linking number is to prove an intermediate result: they both can be calculated from a link diagram by counting positive/negative crossings appropriately. The Gauss map version is straightforward by using the result that degree is the sum of local degrees of the preimages of a point for which the map is a local homeomorphism (then take the crossing points). For the second version of the linking number, you can use Seifert's algorithm to construct a suitable Seifert surface where it is easy to calculate the algebraic intersection number.

For a more direct correspondence, the idea is that given two maps $f_i:S^1\to S^3$, $i=1,2$, with disjoint images $C_1$ and $C_2$, then there is a map $f_1\times f_2:S^1\times S^1\to S^3\times S^3-\Delta$, where $\Delta=\{(x,x):x\in S^3\}$ is the diagonal. Since the knots avoid some point $\infty\in S^3$, this can be given as a map to $\mathbb{R}^3\times\mathbb{R}^3-\Delta$ instead. This space has a map $$g:\mathbb{R}^3\times\mathbb{R}^3-\Delta \to S^2$$ defined by $$g(x,y)=\frac{x-y}{\lVert x-y\rVert}.$$ Notice that the map defined by $(x,y)\mapsto (y,\lVert x-y\rVert, g(x,y))$ gives a homeomorphism $\mathbb{R}^3\times\mathbb{R}^3-\Delta\cong \mathbb{R}^3\times \mathbb{R}_{>0}\times S^2$. Hence, $g$, being the projection onto the third component of this map, is a homotopy equivalence, and as such is essentially unique up to orientation reversal of $S^2$. The composition $g\circ (f_1\times f_2):S^1\times S^1\to S^2$ is the Gauss map. The claim is that $$g_*((f_1\times f_2)_*([S^1\times S^1])) = \operatorname{link}(C_1,C_2) [S^2],$$ where $[S^1\times S^1]$ and $[S^2]$ are fundamental classes, and $\operatorname{link}$ is in the sense of your second linking number. In other words, that $\operatorname{link}(C_1,C_2)=\deg(g\circ(f_1\times f_2))$.

The $S^2$ does not matter so much: we just want to identify $(f_1\times f_1)_*([S^1\times S^1])$ in $H_2(\mathbb{R}^3\times \mathbb{R}^3-\Delta)$. This element is the torus $C_1\times C_2$ as a second homology class.

Let $\Sigma_2$ be as before, then $C_1\times \Sigma_2\subset \mathbb{R}^3\times\mathbb{R}^3$ has $C_1\times C_2$ as its boundary. Consider the long exact sequence $$H_3(\mathbb{R}^3\times \mathbb{R}^3)\to H_3(\mathbb{R}^3\times\mathbb{R}^3,\mathbb{R}^3\times \mathbb{R}^3-\Delta)\to H_2(\mathbb{R}^3\times \mathbb{R}^3-\Delta)\to H_2(\mathbb{R}^3\times \mathbb{R}^3)$$ where the middle map is an isomorphism since $\mathbb{R}^6$ is contractible. Thus, to compute the degree, $C_1\times C_2$ is equivalently $C_1\times \Sigma_2$ as a homology class in the relative $H_3$ group.

The diagonal is homeomorphic to $\mathbb{R}^3$, and it is known that the diagonal of $M\times M$ for a manifold $M$ has a tubular neighborhood diffeomorphic to $TM$. Since $T\mathbb{R}^3=\mathbb{R}^3\times \mathbb{R}^3$, $$H_3(\mathbb{R}^3\times\mathbb{R}^3,\mathbb{R}^3\times \mathbb{R}^3-\Delta) \cong H_3(\mathbb{R}^3\times B^3,\mathbb{R}^3\times S^2)$$ by excision, where $B^3\subset \mathbb{R}^3$ is the unit ball and $S^2=\partial B^3$. Using the long exact sequence, one can see that this group is isomorphic to $\mathbb{Z}$. By some consideration of what $C_1\times \Sigma_2$ looks like when restricted to the tubular neighborhood, we can see that it is a collection of balls, where if $\alpha\subset C_1$ is a small arc and $D\subset\Sigma_2$ is a small disk where $\alpha\cap D$ contains a single point of intersection, $\alpha\times D$ is a ball representing the concatenated local orientations at the intersection, and the sign determines whether this ball is $\pm$ the generator of $H_3(\mathbb{R}^3\times\mathbb{R}^3,\mathbb{R}^3\times \mathbb{R}^3-\Delta)$. Thus, this is counting algebraic intersection number between $C_1$ and $\Sigma_2$, so it is in fact calculating $\operatorname{link}(C_1,C_2)$.

(While we are here, let's understand $g$ a little better. By a suitable change of coordinates, the diagonal in $\mathbb{R}^3\times\mathbb{R}^3$ is diffeomorphic to $\mathbb{R}^3$ in $T\mathbb{R}^3$: at each point of the diagonal, the space orthogonal to the diagonal at that point is a copy of the tangent space of $\mathbb{R}^3$ at that point. So, we can see $\mathbb{R}^3\times\mathbb{R}^3-\Delta$ is $T\mathbb{R}^3-\mathbb{R}^3$ is $\mathbb{R}^3\times(\mathbb{R}^3-0)$. This is homotopy equivalent to $\mathbb{R}^3\times S^2$, which in turn is homotopy equivalent to $S^2$. The $S^2$ gets a standard orientation from the orientation of $\mathbb{R}^3$. A homotopy inverse is any map that sends $S^2$ to the unit tangent vectors of some point within $T\mathbb{R}^3-\mathbb{R}^3$.)

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Your integral that calculates the winding number with the $Z$ axis is indeed calculating a linking number with the $Z$ axis, which is the stereographic projection of an unknot in $S^3$.

You can show this two definitions are equivalent using the Gauss integral definition of the linking number:

Proposition. Suppose $\gamma^1,\gamma^2:S^1\rightarrow\mathbb R^3$ are two oriented smooth embeddings with disjoint images. Then the degree of the map $F:S^1\times S^1\rightarrow S^2$ given by $(s,t)\mapsto (\gamma^1(s)-\gamma^2(t))\cdot||\gamma^1(s)-\gamma^2(t)||^{-1}$ is equal to $$\frac{1}{4\pi}\oint_{\gamma_1}\oint_{\gamma_2}\frac{\mathbf{r}_1-\mathbf r_2}{||\mathbf{r}_1-\mathbf r_2||^3}\cdot (d\mathbf r_1\times d\mathbf r_2).$$

There is a functor from $Top$ to the category of short exact sequences of abelian groups such that for each topological space $X$, we get a short exact sequence $$0\rightarrow Ext (H_{n-1}(X);\mathbb R)\rightarrow H^n(X;\mathbb R)\xrightarrow\beta Hom(H_n(X);\mathbb R)\rightarrow 0,$$ where $\beta$ is given by $\beta([f])([\alpha])=f(\alpha)$. This is a particular version of the universal coefficients theorem.

Recall the $Ext$ functor on the category of pairs of abelian groups. For any abelian group $G$ we have $Ext(\mathbb Z/n\mathbb Z,G)=G/nG$.

Now, if $M$ is a compact manifold, then $Ext(H_{k-1}(M),\mathbb R)=0$ as $\mathbb R$ is a divisible abelian group, for all $k\geq 1$, so we get a natural isomorphism $H^k(M;\mathbb R)\cong Hom(H_k(M);\mathbb R)$ as described above. However, as the De Rham's isomorphism is also natural, we obtain a natural isomorphism $\psi_M: H_{DR}^k(M)\rightarrow Hom(H_k(M);\mathbb R)$, which is the composition of the De Rham's isomorphism $H_{DR}^k(M)\rightarrow H^k(M;\mathbb R)$ and the map $H^k(M;\mathbb R)\rightarrow Hom(H_k(M);\mathbb R)$ just given. This isomorphism is given by $$\psi_M([\omega])([\alpha])=\int_{\alpha}\omega;$$ of course we're using $C^\infty$ chains to compute $H_k(M)$.

If $M$ is orientable and of dimension $n$, there is a canonical generator of $H_n(M)$, which can be obtained from a triangulation of $M$, and is a chain $\sum_{\sigma}(-1)^{i_\sigma}\sigma$, where $\sigma$ goes through the set of $n$-simplices of the triangulation and the signs $(-1)^{i_\sigma}$ can be uniquely obtained from the orientation of $S$ in order to obtain a cycle. We denote this element by $[M]$; this is the so called fundamental class. Then it is clear by this construction that $\int_M\omega=\int_{[M]}\omega$ for all $n$-forms $\omega$ of $M$.

Thus, by using the canonical isomorphism $\psi_M$, we can determine a unique element of $H_{DR}^n(M)$ by taking the only element $[\omega]\in H_{DR}^n(M)$ such that $\psi_M([\omega])=1,$ i.e., $1=\int_{[M]}\omega=\int_M\omega$. Let us denote such a $\omega$ by $\omega_M$; this is unique up to class in $H_{DR}^n(M)$.

Now, if we have two oriented manifolds $M$ and $M´$ of dimension $n$, and a smooth map $f:M\rightarrow M´$, we have, because of the naturality of $\psi_{\_}$, a commutative diagram $\require{AMScd}$ \begin{CD} H^n_{DR}(M')@>\psi_{M'}>> Hom(H_n(M');\mathbb R)\\ @Vf^\ast VV @VVV\\ H_{DR}^n(M)@>\psi_{M}>> Hom(H_n(M);\mathbb R), \end{CD}

where the vertical arrow to the right is the function $g\mapsto g\circ f_\ast$. We have

$$ \begin{split} \int_Mf^\ast\omega_{M'}&=\psi_M(f^\ast([\omega_{M'}]))([M])=\psi_{M'}([\omega_{M'}])\circ f_\#([M])\\ &=\psi_{M'}([\omega_{M'}])(\deg(f)[M'])=\deg(f)\psi_{M'}([\omega_{M'}])([M'])\\ &=\deg(f).(\clubsuit) \end{split} $$

What follows is a modification of an argument given in S. Morita's Geometry of Differntial Forms:

Now, let us consider the following differential form of $\mathbb R^3$: $$\omega:=x_1dx_2\wedge dx_3-x_2dx_1\wedge dx_3+x_3dx_1\wedge dx_2.$$

Then it is clear that $d\omega=3dx_1\wedge dx_2\wedge dx_3$. Let $\omega_0$ be th restriction of $\omega$ to the submanifold $S^2$ of $\mathbb R^3$. We have by Stokes theorem that since $\partial D^3=S^2$ $$\int_{S^2}\omega_0=\int_{D^3} d\omega=3\int_{D^3}dx_1\wedge dx_2\wedge dx_3=4\pi,$$

as the volume of a unitary $3$-ball is $\frac{4}{3}\pi$. Notice $\omega_0$ is a closed form as it is a $2$-form of $S^2$. From this and $(\clubsuit)$ we get $$\deg(F)=\frac{1}{4\pi}\int_{S^1\times S^1}F^\ast\omega_0.$$

Now consider the function $\hat F:S^1\times S^1\rightarrow\mathbb R^3\setminus\{0\}$ given by $(s,t)\mapsto \gamma_1(s)-\gamma_2(t)$, and the function $P:\mathbb R^3\setminus\{0\}\rightarrow S^2$ given by $x\mapsto x/||x||$. Then $P$ is a homotopy equivalence, with homotopic inverse the inclusion $i:S^2\rightarrow \mathbb R^3\setminus\{0\}$. Notice $F=P\circ \hat F$.

Consider the $2$-form of $\mathbb R^3\setminus\{0\}$ given by $$\eta:=\frac{1}{||x||^3}(x_1dx_2\wedge dx_3-x_2dx_1\wedge dx_3+x_3dx_1\wedge dx_2),$$ then it is easy to see that $d\eta=0$. Also, $\eta$ restricted to $S^2$ equals $\omega_0$, i.e., $i^\ast\eta=\omega_0$. We have $$ \begin{split} \deg(F)&=\frac{1}{4\pi}\int_{S^1\times S^1}F^\ast\omega_0=\frac{1}{4\pi}\int_{S^1\times S^1}(P\circ\hat F)^\ast\omega_0\\ &=\frac{1}{4\pi}\int_{S^1\times S^1}(\hat F^\ast\circ P^\ast)\omega_0=\frac{1}{4\pi}\int_{S^1\times S^1}\hat F^\ast\eta \end{split} $$ where the last equality follows from $P^\ast([\omega_0])=[\eta]$ as $i^\ast([\eta])=[\omega_0]$ and $P$ and $i$ are homotopic inverses.

$\gamma^i$ is a vector valued function $(\gamma_1^i,\gamma_2^i,\gamma_3^i)$, for $i=1,2$. We have $\hat F^\ast dx_j={\gamma_1^j} '(s)ds-{\gamma^j_2}'(t)dt$ for $j=1,2,3$. From this it is not hard to prove that $$\deg(f)=\frac{1}{4\pi}\int_{S^1\times S^1}\hat F^\ast\eta=\frac{1}{4\pi}\int_{S^1\times S^1}\frac{\det(\gamma_1'(s),\gamma_2'(t),\gamma_1(s)-\gamma_2(t))}{||\gamma_1(s)-\gamma_2(t)||^3}dsdt.$$

Finally, it is not hard to see this last integral is equal to the integral of the proposition.

In these notes it is shown, in section 1.4, that the definition of the linking number using the integral in our proposition and the definition of the linking number using a link diagram coincide.

Now let us show show your definition coincides with the definition using the degree of the Gauss map.

Suppose we have a smooth embedding $\gamma:S^1\rightarrow \mathbb R^3$. Then it is clear we can build another embedding $\eta:S^1\rightarrow\mathbb R^3$ disjoint from $\gamma$ such that one piece of a diagram of the link $\gamma\cup\eta$ looks like this

$\hskip2in$

Then $[\eta]$ is a generator of the group $H_1(\mathbb R^3\setminus\gamma)$; this is a standard knot theory fact; see for instance Lickorish's book on knots. We can define the linking number of $\gamma$ with another disjoint oriented knot $K$, by simply defining $Lk(\gamma,K)$ as the interger $n$ such that $[K]=n[\eta]$ in $H_1(\mathbb R^3\setminus\gamma)$. In Lickorish book it is proved that the definition using a diagram and this definition coincide, so in particular it coincides with our definition using the integral in the proposition.

Fix a $1$-form $\omega$ of $\mathbb R^3\setminus\gamma$ such that $\psi_{\mathbb R^3\setminus\gamma}([\omega])([\eta])=1$. If $K:S^1\rightarrow\mathbb R^3\setminus\gamma$ is a smooth embedding and $n$ is such that $[K]=n[\eta]$ in $H_1(\mathbb R^3\setminus\gamma)$, then $$\int_K\omega=\psi_{\mathbb R^3\setminus\gamma}([\omega])([K])=\psi_{\mathbb R^3\setminus\gamma}([\omega])(n[\eta])=n.$$

Therefore, your definition coincides with the definition using the first homology group of the complement of a knot, and thus coincides with the definition using the integral of our proposition, and thus in turn it coincides with the definition using the degree of the map $F$.