Direct Image by a Blow up Let $\pi : \widetilde{X} \longrightarrow X$ be the blow up morphism of $X$ a long of $Y \subset X$, with exceptional divisor $E$ and $\text{dim}Y > 0$,   where $X$ and $Y$ are smooth projectives schemes. 
For an answer to a question asked here in the forum (Blowing up and direct image of line bundle), Blowing up and direct image of line bundle), we have: $$\pi_{*}\mathcal{O}_{\widetilde{X}}(-E) = I_{Y/X} \tag{$**$}$$ here $I_{_{Y/X}}$ denotes the ideal sheaf of $Y$.
$(**)$ was obtained through the following exact sequence: $$0 \longrightarrow \mathcal{O}_{\widetilde{X}}(-E) \longrightarrow \mathcal{O}_{\widetilde{X}} \longrightarrow \mathcal{O}_{E} \longrightarrow  0$$
My doubt is: How to find $\pi_{*}\mathcal{O}_{\widetilde{X}}(-nE)$ for $n \geq 0$? What if $n < 0 $?
Thank you.
 A: For $n>1$, the natural guess would be that $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I_{Y/X}^n$, and it's true in this case. For $n<0$, we have $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=\mathcal{O}_X$.
To show these claims, we'll work locally on $X$. Assume $X=\operatorname{Spec} A$ is affine and $Y$ is cut out by the ideal $I_Y=(f_1,\cdots,f_r)$. Then we get a surjection $A^r\to (f_1,\cdots,f_r)$ which turns in to a surjection of the graded algebras $\operatorname{Sym}(A^r)\to \bigoplus_{m\geq 0} I_Y^m$ corresponding to the closed immersion $\widetilde{X}\hookrightarrow \Bbb P(A^r)$. Here, the exceptional divisor $E$ corresponds to the line bundle $\mathcal{O}_{\Bbb P(A^r)}(-1)|_{\widetilde{X}}$.
Next, via the restriction map $\mathcal{O}_{\Bbb P(A^r)}(m) \to \mathcal{O}_{\Bbb P(A^r)}(m)|_{\widetilde{X}}$ and the canonical isomorphism  $\mathcal{O}_{\widetilde{X}}(1)\cong \mathcal{O}_{\widetilde{X}}(-E)$, we see that if $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated, then $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n)\to \pi_*\mathcal{O}_{\widetilde{X}}(-nE)$ will be surjective and thus an isomorphism of line bundles. By identifying $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n) = \operatorname{Sym}^n(A^r)$ and $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I^n$, we see that when the global generation condition is satisfied, then we have an isomorphism $\mathcal{O}_{\widetilde{X}}(-nE)\cong I_Y^n$.
By Serre vanishing, this is always the case for any $X,Y$ assuming that $n>>0$. In our case, the argument from the linked post shows that in fact $\mathcal{O}_{\Bbb P(A^r)}(1)$ is globally generated, so $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated for all $n>0$. So we get our claimed isomorphism $\pi_*\mathcal{O}_\widetilde{X}(-nE)\cong I^n_Y$.
For $n<0$, after tensoring the natural exact sequence $0\to \mathcal{O}_\widetilde{X}(nE) \to \mathcal{O}_X\to\mathcal{K} \to 0$ by $\mathcal{O}_\widetilde{X}(-nE)$ we get the sequence $0\to \mathcal{O}_\widetilde{X}\to \mathcal{O}_\widetilde{X}(-nE)\to \mathcal{K}(-nE)\to 0$ (here $\mathcal{K}$ is the structure sheaf of a thickening of $E$). If we prove that $\pi_*\mathcal{K}(-nE)$ has no global sections, then it's the zero sheaf as $X$ is affine, and this would imply $\mathcal{O}_X=\pi_*\mathcal{O}_\widetilde{X}\to \pi_*\mathcal{O}_\widetilde{X}(-nE)$ is an isomorphism.
As $\mathcal{O}_\widetilde{X}(-nE)\cong \mathcal{O}_\widetilde{X}(n)$, we see that it's a negative line bundle, and after restriction to any projective subvariety of $\Bbb P(A^r)$ it will still be a negative line bundle and therefore have no sections. Taking the fiber $\Bbb P(A^r)_y$ for $y\in Y$, we see that this is a projective variety, so there are no sections of $\mathcal{O}_\widetilde{X}(n)$ in the fiber direction along $E\to Y$. But any global section of $\pi_*\mathcal{K}(-nE)$ would come from such a global section of $\mathcal{O}_\widetilde{X}(n)$ because $\mathcal{O}_\widetilde{X}(n)\to \mathcal{K}(-nE)$ is surjective and would remain this way after restricting to $E$. So there cannot be any global sections and we've shown that $\pi_*\mathcal{O}_\widetilde{X}(-nE)=\mathcal{O}_X$ for $n<0$.
