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I would like to know if the function $\frac{1}{(1+\vert z\vert^2)^2}$ is subharmonic on $\mathbf{C}=\mathbf{P}^1-\{\infty\}$.

Motivation. The Fubini-Study metric on the complex projective line $\mathbf{P}^1$ is given by $\frac{i}{2\pi} \frac{dz \wedge d\overline{z}}{(1+\vert z\vert^2)^2} ;$ this depends on the choice of the coordinate $z$.

I can think of two possiblities. Firstly, just an ugly computation with derivatives. Secondly, an argument using properties of the Fubini-study metric, i.e., it is a smooth positive real $(1,1)$-form.

Question. Is the above function subharmonic?

Question. If yes, then is it possible to prove this without calculating derivatives?

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    What's so bad about taking derivatives? Using $\Delta = 4 \partial \overline{\partial}$ and $|z|^2 = z\bar{z}$ this is a trivial exercise (two lines at most). I get $\Delta \frac{1}{(1+|z|^2)^2} = \frac{16|z|^2 - 8}{(1+|z|^2)^4}$ which doesn't look very positive to me (plug in $|z| \lt \sqrt{2}/2$ and $|z| \gt \sqrt{2}/2$).2011-06-20

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No, this function is not subharmonic on the whole plane, as an easy computation can show (its $\partial \bar \partial$ equals something like $2(2|z|^2-1)/(1+|z|^2)^4$.)

As for the Fubini-Study metric, it can be viewed as the curvature form of the so-called Fubini-Study metric on the line bundle $\mathcal O_{\mathbb P^1}(1)$, its local potential on $\mathbb P^1 - \{\infty\}$ is given by $\varphi(z)= \log (1+|z|^2)$, so that the metrics is $i \partial \bar \partial \varphi$, which is well-defined globally. Its local expression is then as you wrote

$\omega_{\rm FS}=\frac{i}{2\pi} \frac{dz\wedge d\bar z}{(1+|z|^2)^2}$

To sum up, whenever you have a positive (1,1) form $\omega$ on a manifold, then its local potentials (ie the functions $\varphi$ such that $\omega = i\partial \bar \partial \varphi$ locally) are (pluri)subharmonic functions, but you can't say anything about the term in $dz \wedge d\bar z$, except that it is positive (or in higher dimension, the terms in $dz_i \wedge d\bar z_j$ form a positive definite hermitian matrix).

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    @Theo: You're perfectly right, I've edited the answer.2011-06-20
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Subharmonic means http://en.wikipedia.org/wiki/Subharmonic_function the value at the center of a circle is $\le$ the average value around the circle? Try a circle centered at the origin.

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    The computation (made by Theo in the comments, or see my answer) show that your function is subharmonic exactly in the domain \{|z|^2>1/2\}.2011-06-20