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Is the set of real $2n \times 2n$ skew-symmetric matrices having positive Pfaffians path connected?

By definition, the Pfaffian is a polynomial in the entries $a_{ij}$ ($i<j$) such that $Pf(A)^2=\det A$, and $Pf(J_n)=1$, where $$J_n=\begin{pmatrix} 0_n & I_n \\ -I_n & 0n \end{pmatrix}.$$

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  • $\begingroup$ For odd dimension, this intersection is empty. $\endgroup$
    – Asaf Shachar
    Commented Aug 12, 2016 at 12:33
  • $\begingroup$ I really meant even dimension. $\endgroup$
    – zionnn
    Commented Aug 12, 2016 at 12:34
  • $\begingroup$ And also note that for $n=2$ its true. $\endgroup$
    – zionnn
    Commented Aug 12, 2016 at 12:35
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    $\begingroup$ Wait, this seems to be false for $n=2$. $\det \left( \begin{smallmatrix} 0 & x \\ -x & 0 \end{smallmatrix} \right) = x^2 \geq 0$ so all invertible $2 \times 2$ skew symmetric matrices are in $GL_n^+$, and there are two connected components, $x>0$ and $x<0$. A more reasonable conjecture would be that the set of $(2n) \times (2n)$ skew symmetric matrices with positive Pfaffian is path connected. $\endgroup$
    – David E Speyer
    Commented Aug 12, 2016 at 12:41
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    $\begingroup$ Why for $n=2$ is true? The space of $2$-dimensional skew symmetric matrices is a a copy of $\mathbb{R}$ linearly embedded in $\textrm{Mat}(2, \, \mathbb{R})$. The only rank $2$ skew-symmetric matrix with non-positive determinant is the zero matrix, so the intersection with the positive component of $\textrm{GL}(2, \, \mathbb{R})$ is precisely $\mathbb{R} - \{0\}$, which is not path connected. $\endgroup$
    – Francesco Polizzi
    Commented Aug 12, 2016 at 12:42

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The answer is 'yes', because this space is the homogeneous space $\mathrm{GL}^+(2n,\mathbb{R})/\mathrm{Sp}(n,\mathbb{R})$, which is connected.

Here is more detail: Let $J_n$ be the $2n$-by-$2n$ matrix that is $n$ diagonally placed copies of the $2$-by-$2$ matrix $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Then $J_n$ is a skew-symmetric matrix with positive Pfaffian. Any other skew-symmetric $2n$-by-$2n$ matrix $A$ with nonzero determinant can be written in the form $$ A = aJ_na^{T} $$ where $a$ is an element of $\mathrm{GL}(2n,\mathbb{R})$ that is unique up to right multiplication by an element of $$ \mathrm{Sp}(n,\mathbb{R}) = \left\{ a\in\mathrm{GL}(2n,\mathbb{R}) \ |\ \ aJ_na^{T} = J_n\right\}. $$ Finally, note that $\mathrm{Pf}(A) = \det(a)$, so $\mathrm{Pf}(A)>0$ if and only if $a\in\mathrm{GL}^+(2n,\mathbb{R})$. Thus, your space of matrices is identified with the homogeneous space $\mathrm{GL}^+(2n,\mathbb{R})/\mathrm{Sp}(n,\mathbb{R})$, which is connected, since $\mathrm{GL}^+(2n,\mathbb{R})$ is connected.

Remark: Actually, the Pfaffian is not defined as 'the' square root of $\det(A)$, as this is ambiguous. It just happens that the Pfaffian of $A$ is a polynomial in the entries of $A$ whose square is the determinant of $A$ (which is why it is interesting).

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  • $\begingroup$ Great! This is also path-connected because? Sorry for this silly question. I am not a mathematician. Just a physics student. $\endgroup$
    – zionnn
    Commented Aug 12, 2016 at 15:06
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    $\begingroup$ If I understand your question, you are asking whether connected implies path-connected. The answer is yes, because it's a (smooth) manifold (of dimension $2n^2-n$), and a manifold is path-connected if and only if it is connected. $\endgroup$
    – Robert Bryant
    Commented Aug 12, 2016 at 15:36

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