Let $X\subset\mathbb P^n$ be a set of $4$ points and let $I_X\subset\mathcal O_{\mathbb P^n}$ be it's sheaf of ideals. I'd like to compute $H^1(I_X(r))$ for all $r\geqslant0.$
I've tried to consider a long exact cohomology sequence $$\dots\to H^0(\mathcal O_X(r))\to H^1(I_X(r))\to H^1(\mathcal O_{\mathbb P^n}(r))\to H^1(\mathcal O_X(r))\to\dots$$ where $\mathcal O_X=\mathcal O_{\mathbb P^n}/I_X$ but didn't come up with anything worthwhile.
Could you please help me?
From the calculation of $H^i(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r))$, we know that this cohomology group is nonzero iff $i=0$ and $r\geq 0$ or $i=n$ and $r\leq -n-1$. Since a finite collection of points is affine, we know that $H^i(X,\mathcal{F})=0$ for all quasi-coherent $\mathcal{F}$ and all $i>0$. Therefore $H^i(I_X(r))=0$ for all $i \geq 2$, and for $i=0,1$ we are left to examine $$0\to H^0(\Bbb P^n,I_X(r)) \to H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r)) \to H^0(X,\mathcal{O}_X(r)) \to H^1(\Bbb P^n,I_X(r)) \to 0.$$
Since every locally free sheaf on a discrete space is actually free, we see that $\mathcal{O}_X(r)\cong \mathcal{O}_X$ which has a 4-dimensional space of global sections. By the computation of sheaf cohomology of projective space above, we see that $H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r))$ has dimension $\binom{n+r}{n}$. Therefore once we know the rank of the map $H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r))\to H^0(X,\mathcal{O}_X(r))$, we can calculate the dimensions of $H^0(I_X(r))$ and $H^1(I_X(r))$ by rank-nullity.
For $r\geq 3$, I claim this map is surjective: for each point of $X$, we can find a hyperplane of $\Bbb P^n$ which contains it and no other points. Now given a point $x\in X$, a scaling of the product of appropriate powers of the equations for the three hyperplanes containing the other points gives us an element of $H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r))$ which has image one of the standard basis vectors in $H^0(X,\mathcal{O}_X(r))\cong \bigoplus_{x\in X} k$. Therefore when $r\geq 3$, we find that $H^1(I_X(r))=0$ and $\dim H^0(I_X(r))=\binom{n+r}{n}-4$.
When $r=0$, we know that $H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(r))$ is just the constants, and $H^0(\Bbb P^n,I_X(r))$ is the submodule vanishing on all the points hence it's zero. In this case $H^0(\Bbb P^n,I_X(r))=0$ and $\dim H^1(\Bbb P^n,I_X(r))=3$.
When $r=1$ or $r=2$, this is sensitive to the configuration of the points and how big $n$ is. If $r=1$, nonzero elements of $H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(1))$ cut out a hyperplane, so $H^0(I_X(1))$ is nonzero iff all 4 points lie in a hyperplane. In fact, $\dim H^0(I_X(1))$ is the codimension of the smallest linear space containing $X$, and after doing some arithmetic with dimensions, we see that $\dim H^1(I_X(r))=\dim H^0(I_X(1))-n+3$, so $\dim H^1(I_X(1))$ is the difference of the expected dimension of the linear span of $X\subset\Bbb P^n$ and the actual dimension of the linear span of $X\subset\Bbb P^n$. So if the points of $X$ are in general position, $H^1=0$.
When $r=2$, $\dim H^0(\Bbb P^n,\mathcal{O}_{\Bbb P^n}(2))$ measures how many independent quadrics $X$ is on, and the same sort of arithmetic above shows that $\dim H^1$ is the difference between the expected value ($\binom{n+2}{2}-4$) and the actual value. Again, if the points are in general position, $H^1=0$.