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I'm trying to find the degree of the map $f(z) = z^3 - z: \mathbb{C} \rightarrow \mathbb{C}$. The degree is defined as $\deg f = \left( \int_\mathbb{C}\omega \right)^{-1}\int_\mathbb{C}f^*\omega$, where $\omega$ is any differential form on $\mathbb{C}$, e.g. the volume form $dz$. I have problems understanding and finding $f^*$. Can anyone help me?

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  • $\begingroup$ I think $f^*dz=d(z\circ f)=d(z^3-z)=(3z^2-1)dz$. $\endgroup$ – A. Goodier Aug 7 '20 at 13:29
  • $\begingroup$ Thanks! That's very useful. This means that my choice of $\omega = dz$ isn't particularly good since I have to divide with the volume of $\mathbb{C}$. However, I probably want to chose $\omega$ to be an exact form: $\omega = d\eta$, such that $f^*\omega = df(\eta)$. Now the question is which $\eta$ I should use. Any ideas how to think when choosing $\eta$? $\endgroup$ – A.Dunder Aug 7 '20 at 14:06
  • $\begingroup$ Your definition is incorrect – you need additional conditions on $\omega$. The simplest one I can think of is $f$ proper and $\omega$ compactly supported. $\endgroup$ – Mindlack Aug 7 '20 at 14:33
  • $\begingroup$ Yes that makes sense. Could you be so kind to provide an example of an exact compactly supported form on $\mathbb{C}$? $\endgroup$ – A.Dunder Aug 7 '20 at 14:48
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    $\begingroup$ There are of course far easier ways to do this (in terms of counting preimage points of a regular value). This you can do with an integral over a curve (the argument principle, for example). $\endgroup$ – Ted Shifrin Aug 7 '20 at 17:24
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$\frac i2 dz\wedge d\bar z = \frac i2(-2i\, dx\wedge dy) = dx\wedge dy$ is the standard area $2$-form on $\Bbb C$. If you pass to the Riemann sphere $\hat{\Bbb C} = \Bbb CP^1$, the area $2$-form (often called the Kähler form) with integral $1$ is $$\omega = \frac i{2\pi} \dfrac{dz\wedge d\bar z}{(1+|z|^2)^2} = \frac i{2\pi} \partial\bar\partial(\log(1+|z|^2)).$$ It's easy to compute the integral $\displaystyle\int_{\hat{\Bbb C}}\omega = \displaystyle\int_{\Bbb C}\frac i{2\pi} \frac{dz\wedge d\bar z}{(1+|z|^2)^2} = 1$ in polar coordinates. The holomorphic function $g(z)=z^3$ on $\Bbb C$ extends to be a holomorphic function on $\hat C$, and $$g^*\omega = \frac i{2\pi}\dfrac{dg\wedge d\bar g}{(1+|g(z)|^2)^2} = \frac i{2\pi} \frac{9|z|^4\,dz\wedge d\bar z}{(1+|z|^6)^2}.$$ The integral $\displaystyle\int_{\hat{\Bbb C}} g^*\omega$ can likewise be evaluated in polar coordinates, and — surprise! surprise! — you get $3$. And easy homotopy argument then shows that $\int_{\hat{\Bbb C}}f^*\omega = \int_{\hat{\Bbb C}}g^*\omega$. You can also see $\int_{\hat{\Bbb C}} g^*\omega = 3$ by noting that we have a $3$-sheeted covering map on $\hat{\Bbb C} - \{0,\infty\}$.

Alternatively, the degree of the map is gotten by counting preimages of a regular value, with orientations. For a holomorphic map, every regular point appears with a $+1$ orientation, since holomorphic maps are orientation-preserving (note that the determinant of the Jacobian of the map $\Bbb R^2\to\Bbb R^2$ is $|f'(z)|^2$). Now, we can count preimages of a regular value $w$ by evaluating $$\frac1{2\pi i}\int_\gamma \frac{f'(z)\,dz}{f(z)-w} = \frac1{2\pi i}\int_\gamma d\log(f(z)-w)$$ for any curve $\gamma$ that has all those preimages in its interior. Once again, it is easiest to observe that on a large circle you can homotope $f-w$ to $g-w$ without going through $0$, and then the integral just counts the three cube roots of $w$ (for $|w|$ relatively small). This is, of course, geometrically the winding number of the image of the large circle about the origin, and that winding number is locally independent of $w$.

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  • $\begingroup$ Thanks! A few comments: $\\$ 1) How do you find the area form of $\hat{\mathbb{C}}$ knowing the area form of $\mathbb{C}$? $\\$ 2) When is $f^*\omega(z) = \omega(f(z))$ true? Where $\omega$ is some form $\omega(z) = g(z)dz\wedge d\bar{z}$. $\\$ 3) If I want to find the degree of a function $f: \mathbb{R}\rightarrow\mathbb{R}$, should I then pass to a sphere $\hat{\mathbb{R}}$ and consider the volume form of $\hat{\mathbb{R}}$? $\endgroup$ – A.Dunder Aug 9 '20 at 10:07
  • $\begingroup$ You're asking too many unrelated questions. 1) The Kähler form on $\Bbb CP^1$ has nothing to do with the Kähler form on $\Bbb C$. It is, up to scalar multiple, the unique $U(2)$-invariant $(1,1)$-form. 2) Almost never. Write down the definition for pullback. 3) Degree of a function $f\colon\Bbb R\to\Bbb R$ need not be defined. You really need compactness (or proper maps) to even define it. $\hat{\Bbb R}$ is the circle, by the way. $\endgroup$ – Ted Shifrin Aug 9 '20 at 16:11
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Okay I understand it now. I write it here if anyone else is wondering.

The degree of a map $f: M \rightarrow N$ is defined as \begin{equation} \begin{aligned} \deg f = \frac{\int_Mf^*\omega}{\int_N\omega} \ , \end{aligned} \end{equation} where $\omega$ is any compact differential form on $N$ (such that the integrals are finite) and $f^*$ is the pull-back of $f$. E.g. if $N$ has dimension $n$ we can choose \begin{equation} \begin{aligned} \omega = a(x)dx^1\wedge ...\wedge dx^n \ , \end{aligned} \end{equation} where $a(x)$ is any function of $x$ that makes $\omega$ compact. Then the pull-back is given by \begin{equation} \begin{aligned} f^*\omega = a(f(x))df^1\wedge ...\wedge df^n \ . \end{aligned} \end{equation} Here $f(x) = (f^1(x), ..., f^n(x))$ and \begin{equation} \begin{aligned} df^j = \sum_{k = 1}^n\partial_{x^k}f^jdx^k \ . \end{aligned} \end{equation} For example, regarding the function $f(z) = z^3 - z: \mathbb{C}\rightarrow\mathbb{C}$, one can choose \begin{equation} \begin{aligned} \omega = -\frac{e^{-|z|^2}}{2\pi i}dz\wedge\bar{z} = \frac{e^{-x^2 - y^2}}{\pi}dx\wedge dy \quad\Rightarrow\quad \int_{\mathbb{C}}\omega = \int_{\mathbb{R}^2}\frac{e^{-x^2 - y^2}}{\pi}dxdy = \frac{\sqrt{\pi}^2}{\pi} = 1 \ . \end{aligned} \end{equation} Also \begin{equation} \begin{aligned} f^*\omega = -\frac{e^{-|f|^2}}{2\pi i}df\wedge d\bar{f} = -\frac{e^{-|z^3 - z|^2}}{2\pi i}|3z^2 - 1|^2dz\wedge d\bar{z} \ . \end{aligned} \end{equation} As we integrate this over $\mathbb{C}$ we make the variable change $\zeta = (z^3 - z)^{1/3}$ \begin{equation} \begin{aligned} \deg f = \frac{\int_\mathbb{C}f^*\omega}{\int_\mathbb{C}\omega} = -\frac{9}{2\pi i}\int_{\mathbb{C}}e^{-|\zeta|^6}|\zeta|^4dzd\bar{z} \ . \end{aligned} \end{equation} In polar coordinates \begin{equation} \begin{aligned} dz\wedge d\bar{z} = -2irdr\wedge d\theta \ , \quad r > 0 \ , \ \theta\in [0, 2\pi) \ . \end{aligned} \end{equation} Which yields \begin{equation} \begin{aligned} \deg f = \frac{9}{\pi}\int_{0}^{2\pi}\int_0^\infty e^{-r^6}r^5drd\theta = \frac{9}{\pi}\frac{1}{6}2\pi = 3 \ . \end{aligned} \end{equation} In general I feel like choosing $a(x) = e^{-x_1^2 - ... - x_n^2}$ is a good choice since Gaussian integrals are well-behaved. However, for such choice, it might be good to remember the integral \begin{equation} \begin{aligned} \int_0^\infty e^{-r^{2m}}r^ndr = \frac{1}{2m}\Gamma\left(\frac{n + 1}{2m}\right) \ , \end{aligned} \end{equation} which converges for $m > 0$ and $n > -1$.

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There is a much more simpler way applying to such a problem. A holomorphic between Riemann surfaces $f: S \to S'$ induces a homorphism between second homology group $f_*: H_2(S,\mathbb{Z}) \to H_2(S',\mathbb{Z})$. Since every Riemann surface $S$ has a natural orientation $H_2(S,\mathbb{Z}) \cong \mathbb{Z}$ and hence a fundamental class $[S]$, $f_*$ defines an integer $n$ such that $f_*([S])=n[S']$ called the degree. Of course this definition agrees with your original definition.

For a line bundle $L$ on $S$, we define its degree $\mathrm{deg}(L)$ to be the integer under the identification $H^2(S,\mathbb{Z}) \cong \mathbb{Z}$. In other words, $$\mathrm{deg}(L):=\int_M c_1(L)$$ where $c_1$ is the first Chern class which can be thought as $(i/2\pi)\Theta$ for some curvature form $\Theta$.

Let $p$ be a point on $S'$ (in your case, just keep in mind that $S = S'$ is the Riemann sphere because $f$ can be extended to the Riemann sphere) and let $[p],\Theta,f^*([p])=[f^*([p])]$ be the associated line bundle,its curvature form and the pull-back bundle. Therefore, $$\mathrm{deg}f^*([p]) = \int_S f^*\left(\frac{i}{2\pi}\Theta \right) = n\int_{S'}\frac{i}{2\pi}\Theta = n.$$ Consequently, we just need to determine the pull-back divisor $f^*(p)=\sum_{f(q)=p}v_q q$, the degree is just $n=\sum_{f(q)=p}v_q$ for an arbitrary point $p$. The standard local model for a holomorphic map between Riemann surfaces is $z \mapsto z^d$. How could we interpret $d$? That is, the number of solutions of the equations $f(q)=p$ for every $p \in S'$ near $p$. In this case, just choose $p=0$ so $f^{-1}(0)=\left \{0,-1,1 \right\}$ and observe that near each $0,-1,1$ the map is injective, therefore $n=d(0)+d(1)+d(-1)=1+1+1=3$.

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