Consider $\mathcal{M}_{0,n}$, the moduli space of genus zero curves with $n$-punctures.

Using a combination of the Kodaira-Spencer map, Riemann-Roch, and Serre Duality, I have calculated the dimension of $\mathcal{M}_{0,n}$ to be $n-3$.

Using such heavy theorems, I have lost intuition as to why adding a puncture increases the dimension of the moduli space.

Very roughly, using the interpretation of $\mathcal{M}_{0,n}$ as the configuration space of projectively equivalent $n$-tupels points in $\mathbb{P}^1$, I imagine that the configuration class becomes "finer" as we increase the number of punctures. However, my reasoning isn't all that satisfying.

Why is the dimension of the moduli space increasing by exactly one for each puncture?

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    $\begingroup$ The dimension is $n - 3$ because the $n$ is for the $n$-tuple of points in $\mathbb{P}^1$ and the $-3$ is for quotienting by the action of $PGL_2$. Every time $n$ goes up you add one more point so one more copy of $\mathbb{P}^1$; the intuition is clear. $\endgroup$ – Qiaochu Yuan Jan 15 '18 at 5:34
  • $\begingroup$ Thank you. I have never thought it about it like this but very simple now that you say it. $\endgroup$ – Yuugi Jan 15 '18 at 20:34

Let me interpret $\mathcal{M}_{0,n}$ as the moduli space of genus zero curves with $n$ marked points, as opposed to punctures. This is just a psychological thing and should not make a difference to what follows.

That $\dim(\mathcal{M}_{0,n}) = 0$ for $n = 0,1,2,3$ is very classical: this amounts to saying that all genus zero curves (say over an algebraically closed field) are isomorphic to $\mathbf{P}^1$, and that given two pairs $(p_i)$ and $(q_i)$ of ordered tuples consisting of at most $3$ points, there is an automorphism of $\mathbf{P}^1$ sending the $(p_i)$ to the $(q_i)$. This automorphism can be constructed directly once you choose coordinates on $\mathbf{P}^1$.

Okay, now suppose that $n \geq 4$. A point of $\mathcal{M}_{0,n}$ is to be thought of as $\mathbf{P}^1$ together with $n$-points $p_1,\ldots,p_n$, all of this up to automorphisms of $\mathbf{P}^1$. As discussed in the previous paragraph, there is a unique automorphism of $\mathbf{P}^1$ that sends the first three points $p_1,p_2,p_3$ to $0,1,\infty$, in that order. So, spending this piece of freedom, a point of $\mathcal{M}_{0,n}$ can always be represented by $\mathbf{P}^1$ together with the $n$-points $0,1,\infty,p_4,\ldots,p_n$, where now $p_4,\ldots,p_n$ are some $n - 3$ distinct points of $\mathbf{P}^1$ different from $0,1,\infty$. In fact, by uniqueness of the automorphism sending $p_1,p_2,p_3$ to $0,1,\infty$, we see that every point of $\mathcal{M}_{0,n}$ has a unique representative of the form $$ (\mathbf{P}^1, 0,1,\infty,p_4,\ldots,p_n) \quad p_4,\ldots,p_n \in \mathbf{P}^1 \setminus \{0,1,\infty\}\;\text{distinct.} $$ So now we can compute dimensions by figuring out what freedom we have left: the only things that can move are the $n - 3$ points $p_4,\ldots,p_n$. Each $p_i$ is allowed to take all but finitely many values in $\mathbf{P}^1$. Since removing any finite number of points in $\mathbf{P}^1$ still gives a $1$-dimensional object, we see that $\dim(\mathcal{M}_{0,n}) = n - 3$, as you computed.

  • $\begingroup$ This is also a great explanation. Thank you so much, I am happy I asked this! $\endgroup$ – Yuugi Jan 15 '18 at 20:50

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