For a given instanton bundle $E$ there is a STB $S$ associated with $E$, which is stable ([AO], Theorem 2.8) and unique (ibid., Prop. 2.17). It is easy to prove that the map $\pi: M \to N$ defined by $\pi([E]) = [S]$ is algebraic, moreover $\pi$ is dominant by [ST]. If $m = [E] \in M$, the fiber $\pi^{-1}(\pi(m))$ is a Zariski open subset of the grassmannian of planes in the vector space $H^0(\mathbb{P}^{2n+1}, S^*(1))$, where $\pi(m) = [S]$; by the Theorem 3.14 of [AO], $h^0(\mathbb{P}^{2n+1}, S^*(1)) = 2n+2$, hence $\dim \pi^{-1}(\pi(m)) = 4n$.
In order to prove that $M$ is irreducible, we suppose by contradiction that there are at least two irreducible components $M_0$ and $M_1$ of $M$. Then $M_0 \cap M_1 = \emptyset$ ($M$ is smooth), $\pi(M_0)$ and $\pi(M_1)$ are constructible subset of $N$ by Chevalley's theorem. Looking at the dimensions of $M_0, M_1, N$ and the fibers of $\pi$ we conclude that both $\pi(M_0)$ and $\pi(M_1)$ must contain an open subset of $N$, which implies $\pi(M_0) \cap \pi(M_1) \neq \emptyset$ by the irreducibility of $N$. This is a contradiction because the fibers of $\pi$ are connected. □
For $n \ge 2$ and $k \ge 3$, it is no longer true that $\mathrm{MI}_{p^{2n+1}}(k)$ is smooth. In fact on $\mathbb{P}^5$ we have:
Theorem 3.3. The space $\mathrm{MI}{\mathbb{P}^5}(k)$ is singular for $k=3,4$. To be more precise, the irreducible component $M_0(k)$ of $\mathrm{MI}{\mathbb{P}^5}(k)$ containing the special instanton bundles is generically reduced of dimension $54(k=3)$ or $65(k=4)$, and $\mathrm{MI}_{\mathbb{P}^5}(k)$ is singular at the points corresponding to special symplectic instanton bundles.
Proof. Let $E_0$ be the special instanton bundle on $\mathbb{P}^5$ of the Example 2.2($k=3$) or of the Example 2.5($k=4$). Then $h^2(E_0 \otimes E_0^*) = 0$ and $M_0(k)$ is smooth at the point corresponding to $E_0$, of dimension $h^1(E_0 \otimes E_0^*) = 54(k=3)$ or $65(k=4)$. In particular, $M_0(k)$ is generically reduced. If $E_1$ is a special symplectic instanton bundle on $\mathbb{P}^5$, the computations in 2.1 and 2.3 show that $h^2(E_1 \otimes E_1^*) = 3(k=3)$ or $12(k=4)$, and $h^1(E_1 \otimes E_1^*) = 57$ or $77$ respectively. Hence $\mathrm{MI}_{\mathbb{P}^5}(k)$ is singular at $E_1$ for $k=3$ and $4$. □
Remark 3.4. It is natural to conjecture that $\mathrm{MI}_{\mathbb{P}^{2n+1}}(k)$ is singular for all $n \ge 2$ and $k \ge 3$.
Theorem 3.5. Let $E$ be an instanton bundle on $\mathbb{P}^{2n+1}$ with $c_2(E) = k$. Then
Proof. The result is obvious for $t \le -2$. It is sufficient to prove $h^1(S^*(t)) = 0$ for $t \ge k-1$. We have