I'm reading The ADHM construction of Yang-Mills instantons by Simon Donaldson.
A theorem in the section 4 says:
Let $E$ be a rank $r$ holomorphic bundle over $\mathbb{CP}^3$ with $c_2 = k$. Suppose that $E$ is trivial on some line and satisfies the vanishing conditions $$\mathrm{H}^0(E)=0,\quad \mathrm{H}^0(E^*)=0, \quad \mathrm{H}^1(E(-2))=0,\quad \mathrm{H}^1(E^*(-2))=0$$ Then $E$ arises from a monad with $\dim U = \dim W = k$ and $\dim V=2k+r$. The monad is unique up to the action of $\mathrm{GL}(U) \times \mathrm{GL}(V) \times \mathrm{GL}(W)$.
Let me explain the last sentence more. Let $U$, $V$ and $W$ denote any complex vector space. We denote $U$ as a trivial vector bundle on $\mathbb{CP}^3$ by $\underline{U}$, and similarly define $\underline{V}$ and $\underline{W}$. Also, we denote $\underline{U}\otimes \mathcal{O}(-1)$ by $\underline{U}(-1)$, $\underline{V} \otimes \mathcal{O}(1)$ by $\underline{V}(1)$, and so on.
A monad $\underline{U}(-1) \xrightarrow{\alpha} \underline{V} \xrightarrow{\beta} \underline{W}(1)$ is a chain of three bundles satisfying three conditions: $\alpha$ is an injective bundle map, $\beta$ is a surjective bundle map and $\beta \circ \alpha = 0$. In this setting, $E$ is said to be arised from the monad when $E\simeq \ker \beta \,/\, \mathrm{im} \alpha$.
The proof of this theorem is given in the section $5$ using the Beilinson spectral sequence. As a result, $U:=\mathrm{H}^2(E(-3))$, $V:=\mathrm{H}^2(E\otimes \Omega^2)$ and $W:= \mathrm{H}^2(E\otimes \Omega^1(-1))$ work. I worked on some spectral sequences and I reached this point.
However, I can't find any reason for $\dim U = \dim W$, or equivalently $\dim \mathrm{H}^2(E(-3)) = \dim \mathrm{H}^2(E \otimes \Omega^1(-1))$. Why this necessarliy holds?
Thank you.
Although it is quite complicated, I found my own reason somehow.
Let $\mathbb{CP}^3 = \mathrm{Proj} \,\,\mathbb{C}[x_0, x_1, x_2, x_3]$.
Without loss of generality, assume that $E$ is trivial on the line $V(x_0, x_1) \simeq \mathbb{CP}^1$.
Step 1
Consider a short exact sequence $0\rightarrow E(-3)\xrightarrow{\times x_0} E(-2) \rightarrow E(-2)|_{V(x_0)} \rightarrow 0$.
We get a long exact sequence:
$$\cdots \rightarrow \mathrm{H}^1(E(-2)) \rightarrow \mathrm{H}^1(E(-2)|_{V(x_0)}) \rightarrow \mathrm{H}^2(E(-3))\rightarrow \mathrm{H}^2(E(-2))\rightarrow \cdots$$
The condition $\mathrm{H}^1(E(-2)) = 0$ is given.
Also, by the Serre duality,
$$0 = \mathrm{H}^1(E^*(-2)) = \mathrm{H}^1(E^* (2) \otimes \mathcal{O}(-4)) = \mathrm{H}^1((E(-2))^* \otimes \Omega^3) \simeq \mathrm{H}^2(E(-2))^*$$
and $\mathrm{H}^2(E(-2)) = 0$ holds.
Therefore $\mathrm{H}^2(E(-3)) \simeq \mathrm{H}^1(E(-2)|_{V(x_0)})$.
Step 2
Think of the Euler sequence $0 \rightarrow \Omega^1 \rightarrow \mathcal{O}(-1)^{\oplus 4} \rightarrow \mathcal{O}\rightarrow 0$.
Since $E(-1)$ is locally free, we get a short exact sequence $$0 \rightarrow E \otimes \Omega^1 (-1) \rightarrow E(-2)^{\oplus 4} \rightarrow E(-1) \rightarrow 0$$
and its long exact sequence $$ \cdots \rightarrow \mathrm{H}^1(E(-2)^{\oplus 4}) \rightarrow \mathrm{H}^1(E(-1)) \rightarrow \mathrm{H}^2 (E\otimes \Omega^1(-1)) \rightarrow \mathrm{H}^2(E(-2)^{\oplus 4}) \rightarrow \cdots$$
We know $\mathrm{H}^1(E(-2)^{\oplus 4}) = \mathrm{H}^2(E(-2))^{\oplus 4} = 0$.
Similarly $\mathrm{H}^2(E(-2)^{\oplus 4}) = 0$ and we get $\mathrm{H}^2(E\otimes \Omega^1(-1)) \simeq \mathrm{H}^1(E(-1))$.
Step 3
Consider a short exact sequence $0\rightarrow E(-2) \xrightarrow{\times x_0} E(-1) \rightarrow E(-1)|_{V(x_0)} \rightarrow 0$.
It yields a long exact sequence $$\cdots \rightarrow \mathrm{H}^1(E(-2)) \rightarrow \mathrm{H}^1(E(-1)) \rightarrow \mathrm{H}^1(E(-1)|_{V(x_0)}) \rightarrow \mathrm{H}^2(E(-2)) \rightarrow \cdots$$
Again, $\mathrm{H}^1(E(-2))$ and $\mathrm{H}^2(E(-2))$ vanish.
Therefore $\mathrm{H}^1(E(-1)) \simeq \mathrm{H}^1(E(-1)|_{V(x_0)})$.
Step 4
Consider a short exact sequence $0 \rightarrow E(-2)|_{V(x_0)} \xrightarrow{\times x_1} E(-1)|_{V(x_0)} \rightarrow E(-1)|_{V(x_0, x_1)} \rightarrow 0$.
Its long exact sequence is: $$\cdots \rightarrow \mathrm{H}^0 (E(-1)|_{V(x_0, x_1)}) \rightarrow \mathrm{H}^1(E(-2)|_{V(x_0)}) \rightarrow \mathrm{H}^1(E(-1)|_{V(x_0)}) \rightarrow \mathrm{H}^1(E(-1)|_{V(x_0, x_1)}) \rightarrow \cdots$$
Since we assumed $E|_{V(x_0, x_1)}$ is trivial, $E(-1)|_{V(x_0, x_1)} \simeq \mathcal{O}(-1)^{\mathrm{rank} E}$.
But $\mathrm{H}^\bullet (\mathcal{O}(-1)|_{V(x_0, x_1)})=0$ because $V(x_0, x_1) \simeq \mathbb{CP}^1$ and $\mathrm{H}^\bullet(\mathbb{CP}^1,\mathcal{O}(-1))=0$.
Therefore $\mathrm{H}^\bullet(E(-1)|_{V(x_0, x_1)}) = 0$ and the long exact sequence implies $$\mathrm{H}^1(E(-2)|_{V(x_0)}) \simeq \mathrm{H}^1(E(-1)|_{V(x_0)})$$
Conclusion
Finally we get $$\mathrm{H}^2 (E\otimes \Omega^1(-1)) \stackrel{\text{step 2}}{\simeq} \mathrm{H}^1(E(-1)) \stackrel{\text{step 3}}{\simeq} \mathrm{H}^1(E(-1)|_{V(x_0)}) \stackrel{\text{step 4}}{\simeq} \mathrm{H}^1(E(-2)|_{V(x_0)}) \stackrel{\text{step 1}}{\simeq} \mathrm{H}^2(E(-3))$$
and this completes the proof.