6 Completed the argument in the semi-simple case
for for any $m:V\to V$. When $\dim_k(V) = 2n$, let $\mathsf{K}(V) = \Lambda^{2n}(V^\ast)$ denote the $1$-dimensional vector space consisting of $2n$-multilinear alternating functions on $V$. There exists a canonical polynomial mapping $\mathrm{Pf}:\mathsf{A}(V)\to \mathsf{K}(V)$ of degree $n$ that satisfies $\alpha^n = n!\ \mathrm{Pf}(\alpha)$ (a property that defines $\mathrm{Pf}$ when $n$ is less than the characteristic of $k$). This Pfaffian vanishes if and only if $\alpha$ is degenerate, and it satisfies $\mathrm{Pf}\bigl(m^\ast(\alpha)\bigr) = \det(m)\ \mathrm{Pf}(\alpha)$.

Anyway, back to the question: One approach is to look at the orbits of $Sp(\alpha)$ acting on ${\frak{s}}(\alpha)$ and seeing see what their squares look like. This may be easier because the adjoint orbits of $Sp(\alpha)$ on its Lie algebra have been much studied.

As an example, suppose that $k$ is algebraically closed and (for my comfort) that it has characteristic zero. Say that a pair of nondegenerate alternating $2$-forms $(\alpha_0,\alpha)$ on $k^{2n}$ is generic if the $n$ roots of the equation $\textrm{Pf}(\alpha - \lambda\ \alpha_0) = 0$ are all distinct. Then one can prove (see below) that a basis of $1$-forms on $k^{2n}$ exists so thatThus, the problem uncouples into $n$ separate problems that are each trivially solvable. Thus, so, the problem is solvable for the generic pair in this case.

To prove surjectivity for all $n$, one needs may need to understand the orbits of $Sp(\alpha)$ acting on ${\frak{a}}(\alpha)$. I think that this is a classical problem (I'm not an algebraist, so I'm not completely sure), so maybe it's time to look at the literature. The classification of the possible $Sp(\alpha)$-orbit types in ${\frak{a}}(\alpha)$ gets more complicated as $n$ increases, so maybe some other approach needs to be tried. One would expect the orbits of $Sp(\alpha)$ in ${\frak{a}}(\alpha)$ to be somewhat simpler than the orbits of $Sp(\alpha)$ in ${\frak{s}}(\alpha)$, just because the dimension is lower. However, I note that the rings of $Sp(\alpha)$-invariant polynomials on each of the vector spaces ${\frak{s}}(\alpha)$ and ${\frak{a}}(\alpha)$ are each free polynomial rings on $n$ generators, so it may be that the complexity of the orbit structures are (at least roughly) comparable in the two cases.

The uncoupling step: In the general , note thatcase, for $a\in{\frak{a}}(\alpha_0)$, one has $\textrm{Pf}(a^\flat - \lambda\ \alpha_0) = p_a(\lambda)\ \textrm{Pf}(\alpha_0)$, where $\det(a - \lambda I) = p_a(\lambda)^2$. Letting $f_1(\lambda),\ldots,f_k(\lambda)$ denote the distinct irreducible factors of $p_a(\lambda)$, one hasThere is a direct sum decomposition $V = V_1\oplus\cdots\oplus V_k$ into the corresponding generalized eigenspaces of $a$, and one sees without difficulty (using the fact that identity $\alpha_0(ax,y) = \alpha_0(x,ay)$) that $\alpha_0(V_i,V_j) = 0$ for $i\not=j$. Moreover, any solution $s\in{\frak{s}}(\alpha_0)$ of $a = \pm s^2$ must commute with $a$ and therefore preserve its generalized eigenspaces. Thus, generalizing the 'uncoupled' situation given in the first example, one sees that the problem reduces to the case in which $p_a(\lambda)$ is a power of a single irreducible polynomial.

Unfortunately, it turns out that, even in this uncoupled case, the minimal polynomial of $a$ can fail to be irreducible (i.e., $a$ need not be semi-simple), and I do not know a simple way to handle all of these cases. When $n=2$, this can be handled 'by hand', but even for $n=3$, it seems to be a little tricky (although, I think that I have correctly handled the cases there and shown surjectivity in that case as well).

The uncoupled semi-simple case: Here is how one can complete the proof of solvability in the uncoupled, semi-simple case, i.e., when the minimal polynomial of $a$ is irreducible. Let $p(\lambda)=0$ be the (irreducible) minimal polynomial of $a$, say of degree $m$. We can assume that $p(0)\not=0$, since, otherwise, $a=0$, and there is nothing to prove (i.e., one can just take $s=0$, and the problem is solved). Let $K\subset End(V)$ denote the field generated by $a$, so that $[K:k]=m$.

Now, for any nonzero $x\in V$, one has $\alpha_0(K{\cdot}x, K{\cdot}x) = 0$, since $\alpha_0(a^ix,a^jx)=0$ for all $i$ and $j$. If one chooses $x,y\in V$ such that $\alpha_0(x,y)\not=0$, then it is easy to see (using $p(0)\not=0$) that $\alpha_0$ is nondegenerate on the $a$-invariant subspace $W = K{\cdot}x\oplus K{\cdot}y$. One then can write $V = W\oplus W^\perp$ (where the $\perp$ is taken with respect to $\alpha_0$) and see that the problem uncouples into separate problems on $W$ and $W^\perp$. By induction, it then suffices to consider the case $W = V$ (and, hence, $n=m$).

In this case, the general element of $V = W = K{\cdot}x\oplus K{\cdot}y$ can be written uniquely in the form $z = f_1(a)x + f_2(a)y$ where $f_1$ and $f_2$ are polynomials (with coefficients in $k$) of degree at most $m{-}1$. Let $q(a)$ and $r(a)$ be any polynomials in $a$ (i.e., elements of $K$) and define a map $s:V\to V$ by s\bigl(f_1(a)x + f_2(a)y\bigr) = f_1(a)q(a)y + f_2(a)r(a)x.$$One checks that s\in{\frak{s}}(\alpha_0) and notes that, by construction, one has the identity s^2 = q(a)r(a). By setting q(a) = a and r(a) = \pm1, one sees that one may arrange s^2 = \pm a. This s solves the problem. 5 added information about the nilpotent case An element \beta\in \mathsf{B}(V) is nondegenerate if, for each x\not=0 in V, there exists a y\in V such that \beta(x,y)\not=0. If \alpha\in \mathsf{A}(V) is nondegenerate, then the dimension of V over k must be even. Conversely, if the dimension of V over k is even, then there exists a nondegenerate \alpha\in\mathsf{A}(V), and, moreover, any other nondegenerate \overline\alpha\in\mathsf{A}(V) is of the form \overline\alpha = m^\ast(\alpha) for some m\in GL(V), where, by definition, \overline\alpha(x,y) bigl(m^\ast(\alpha)\bigr)(x,y) := \alpha(mx,my)for some for any m\in GL(V)m:V\to V. When \dim_k(V) = 2n, let \mathsf{K}(V) = \Lambda^{2n}(V^\ast) denote the 1-dimensional vector space consisting of 2n-multilinear alternating functions on V. There exists a canonical polynomial mapping \mathrm{Pf}:\mathsf{A}(V)\to \mathsf{K}(V) of degree n that satisfies \alpha^n = n!\ \mathrm{Pf}(\alpha) (a property that defines \mathrm{Pf} when n is less than the characteristic of k). This Pfaffian vanishes if and only if \alpha is degenerate, and it satisfies \mathrm{Pf}\bigl(m^\ast(\alpha)\bigr) = \det(m)\ \mathrm{Pf}(\alpha). When \alpha is nondegenerate, let Sp(\alpha)\subset GL(V) denote the subgroup consisting of those m\in GL(V) such that \alpha(mx,my)=\alpha(x,y) for all x,y\in V. Define two subspaces {\frak{s}}(\alpha)\subset \mathrm{End}(V) \simeq V\otimes V^\ast and {\frak{a}}(\alpha)\subset \mathrm{End}(V), by saying that s\in{\frak{s}}(\alpha) if s^\flat(x,y) := \alpha(x,sy) is symmetric, while a\in{\frak{a}}(\alpha) if a^\flat(x,y) := \alpha(x,ay) is alternating. Note that {\frak{s}}(\alpha) is a subalgebra of V\otimes V^* under the commutator bracket; in fact, it is the Lie algebra of Sp(\alpha). The subspaces {\frak{s}}(\alpha) and {\frak{a}}(\alpha) are invariant under conjugation by elements of Sp(\alpha). Here, then, is the question: What is the image of S? (The OP is actually asking whether the image of S contains the nondegenerate invertible elements of \mathsf{A}(V).){\frak{a}}(\alpha).) As an example, suppose that k is algebraically closed and (for my comfort) that it has characteristic zero. Say that a pair of nondegenerate alternating 2-forms (\alpha_0,\alpha) on k^{2n} is generic if the n roots of the equation (\alpha \textrm{Pf}(\alpha - \lambda\ \alpha_0)^n alpha_0) = 0 are all distinct. Then one can prove that a basis of 1-forms on k^{2n} exists so that To do it prove surjectivity for all n, you need one needs to understand the orbits of Sp(\alpha) acting on {\frak{a}}(\alpha). I think that this is a classical problem (I'm not an algebraist, so I'm not completely sure), so maybe it's time to look at the literature. The classification of the possible Sp(\alpha)-orbit types in {\frak{a}}(\alpha) gets more complicated as n increases, so maybe some other approach needs to be tried. One would expect the orbits of Sp(\alpha) in {\frak{a}}(\alpha) to be somewhat simpler than the orbits of Sp(\alpha) in {\frak{s}}(\alpha), just because the dimension is lower. However, I note that the rings of Sp(\alpha)-invariant polynomials on each of the vector spaces {\frak{s}}(\alpha) and {\frak{a}}(\alpha) are each free polynomial rings on n generators, so it may be that the complexity of the orbit structures are (at least roughly) comparable in the two cases. In general, note that, for a\in{\frak{a}}(\alpha_0), one has \textrm{Pf}(a^\flat - \lambda\ \alpha_0) = p_a(\lambda)\ \textrm{Pf}(\alpha_0), where \det(a - \lambda I) = p_a(\lambda)^2. Letting f_1(\lambda),\ldots,f_k(\lambda) denote the distinct irreducible factors of p_a(\lambda), one hasp_a(\lambda) = f_1(\lambda)^{d_1}\cdots f_k(\lambda)^{d_k}.There is a direct sum decomposition V = V_1\oplus\cdots\oplus V_k into the corresponding generalized eigenspaces of a, and one sees without difficulty (using the fact that \alpha_0(ax,y) = \alpha_0(x,ay)) that \alpha_0(V_i,V_j) = 0 for i\not=j. Moreover, any solution s\in{\frak{s}}(\alpha_0) of a = \pm s^2 must commute with a and therefore preserve its generalized eigenspaces. Thus, generalizing the 'uncoupled' situation given in the first example, one sees that the problem reduces to the case in which p_a(\lambda) is a power of a single irreducible polynomial. Unfortunately, it turns out that, even in this uncoupled case, the minimal polynomial of a can fail to be irreducible (i.e., a need not be semi-simple), and I do not know a simple way to handle all of these cases. When n=2, this can be handled 'by hand', but even for n=3, it seems to be a little tricky (although, I think that I have correctly handled the cases there and shown surjectivity in that case as well). 4 Fixed grammar and added some more information When \alpha is nondegenerate, let Sp(\alpha)\subset GL(V) denote the subgroup consisting of those m\in GL(V) such that \alpha(mx,my)=\alpha(x,y) for all x,y\in V. We can define Define two subspaces {\frak{s}}(\alpha)\subset \mathrm{End}(V) = \simeq V\otimes V^\ast and {\frak{a}}(\alpha)\subset \mathrm{End}(V) = V\otimes V^\ast, mathrm{End}(V), by saying that s\in{\frak{s}}(\alpha) if \beta_s(x,y) s^\flat(x,y) := \alpha(x,sy) is symmetric, while a\in{\frak{a}}(\alpha) if \beta_a(x,y) a^\flat(x,y) := \alpha(x,ay) is alternating. Note that {\frak{s}}(\alpha) is a subalgebra of V\otimes V^* under the commutator bracket, ; in fact, it is the Lie algebra of Sp(\alpha). The subspaces {\frak{s}}(\alpha) and {\frak{a}}(\alpha) are invariant under conjugation by elements of Sp(\alpha). What is interesting is that Now, there is a natural map S:{\frak{s}}(\alpha)\to {\frak{a}}(\alpha), given by S(s) = s^2. In other words, if \alpha(x,sy) is symmetric, then$$\alpha(x,s^2x) \alpha(x,s^2x) = \alpha(sx,sx) = 00$,$$Note that the dimension of${\frak{s}}(\alpha)$is$2n^2{+}n$, while the dimension of${\frak{a}}(\alpha)$is$2n^2{-}n$, so it's conceivable that$S$is actually surjective.Also, the map$S$is$Sp(\alpha)$-equivariant, so it's really a question about that can be studied by looking at the orbits of this group acting on${\frak{a}}(\alpha)$. Remark: It took me a while to recognize that this is what is going on because the question, as asked, sneaks in an extraneous quadratic form that breaks the symplectic symmetry. We were given a `A 'reference' alternating form on$k^{2n}$has been specified by the formula$\alpha_0(x,y) = x^TJy$for$x,y\in k^{2n}$. Note that the matrix$J$satisfies$J^2 = -I$, an identity that has no meaning for an alternating form. The only way one can interpret an alternating form as a linear transformation (so that squaring makes sense) is to have some other way of identifying$V$with$V^\ast$. Of course, this is supplied by the linear map$x\mapsto x^T$in the formula. In other words, a (symmetric) bilinear form$\beta(x,y) = x^Ty$has been introduced into the picture, and it breaks the symplectic symmetry. Anyway, writing$\alpha(x,y) = x^TAy$and asking whether one can write$A = MJM$for$M$symmetric can be interpreted re-interpreted as follows: Note that$M=Js$where$s\in{\frak{s}}(\alpha_0)$and that$\alpha(x,y) = x^TAy = x^TJJ^{-1}Ay = \alpha_0(x,J^{-1}Ay) = \alpha_0(x,ay)$where$a = J^{-1}A$lies in${\frak{a}}(\alpha_0)$. Putting this together says that we want so showing that the desired equation$A = MJM$can be solved is equivalent to showing that the equation$a = -s^2$. s^2$ for a given $a\in{\frak{a}}(\alpha_0)$ can be solved for some $s\in{\frak{s}}(\alpha_0)$. (You'll note that it's It's off by a minus sign, but that's OK because we want the goal is to characterize the image , of $S$, so characterizing its negative is just as good.)

Anyway, back to the question: Now it's a question of looking One approach is to look at the orbits of $Sp(\alpha)$ acting on ${\frak{s}}(\alpha)$ and seeing what their squares look like. This is may be easier because we know something about the adjoint orbits of $Sp(\alpha)$ on its Lie algebra , and Lie theory and other things can be brought to bearhave been much studied.

For

As an example, if suppose that $k$ is algebraically closed and , say (for my comfort), of comfort) that it has characteristic zero, say . Say that a pair of nondegenerate alternating $2$-forms $\alpha$ and $\alpha_0$ (\alpha_0,\alpha)$on$k^{2n}$are is generic if the$n$roots of the equation$(\alpha - \lambda\ \alpha_0)^n = 0$are all distinct. Then one can easily prove that a basis of$1$-forms on$k^{2n}$exists so that Also, one can now, by hand As another example, do in the case$n=2$n=2$, for an arbitrary field (even one of characteristic not $2$ 2$), one can, by hand, classify the pairs$(\alpha_0,\alpha)$with$\alpha_0$nondegenerate and show that , again, it$S:{\frak{s}}(\alpha_0)\to{\frak{a}}(\alpha_0)$is always solvable, even in the nongeneric casessurjective. (I'll put this in explicitly the details if someone asks.) To do it for all$n$, you need to understand the orbits of$Sp(\alpha)$acting on${\frak{a}}(\alpha)$. I think that this is a classical problem (I'm not an algebraist, so I'm not completely sure), so maybe it's time to look at the literature. The classification of the possible$Sp(\alpha)$-orbit types in${\frak{a}}(\alpha)$gets more complicated as$n$increases, so maybe some other approach needs to be tried. One would expect the orbits of$Sp(\alpha)$in${\frak{a}}(\alpha)$to be somewhat simpler than the orbits of$Sp(\alpha)$in${\frak{s}}(\alpha)$, just because the dimension is lower. However, I note that the rings of$Sp(\alpha)$-invariant polynomials on each of the vector spaces${\frak{s}}(\alpha)$and${\frak{a}}(\alpha)$are each free polynomial rings on$n\$ generators, so it may be that the complexity of the orbit structures are (at least roughly) comparable in the two cases.

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