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Exercise 2002Topology › Topology of Mn(ℝ) › Applications and the case n = 2M1Added 6 Nov 2026

Invertible skew-symmetric matrices have two components

#Pfaffian#connected components#skew-symmetric matrices

Statement

Let nn be even and let S={K∈Skewn:det⁡K≠0}\mathcal S=\{K\in\mathrm{Skew}_n:\det K\ne0\}. Show that S\mathcal S has exactly two connected components, distinguished by the sign of the Pfaffian Pf⁡(K)\operatorname{Pf}(K).

Hint

Every nondegenerate skew matrix has the form K=PJPTK=PJP^T with P∈GLn(R)P\in GL_n(\mathbb{R}) and JJ the standard block matrix, and Pf⁡(PJPT)=det⁡P⋅Pf⁡(J)\operatorname{Pf}(PJP^T)=\det P\cdot\operatorname{Pf}(J).

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Solution

Facts used. The Pfaffian is a polynomial in the entries of KK with Pf⁡(K)2=det⁡K\operatorname{Pf}(K)^2=\det K and Pf⁡(PKPT)=det⁡(P)Pf⁡(K)\operatorname{Pf}(PKP^T)=\det(P)\operatorname{Pf}(K). For J=⨁j=1n/2(01−10)J=\bigoplus_{j=1}^{n/2}\begin{pmatrix}0&1\\-1&0\end{pmatrix} one has Pf⁡(J)=1\operatorname{Pf}(J)=1. We also use that GLn+(R)GL_n^+(\mathbb{R}) and GLn−(R)GL_n^-(\mathbb{R}) are connected (standard).

Step 1: the Pfaffian separates. Since det⁡K=Pf⁡(K)2\det K=\operatorname{Pf}(K)^2, K∈SK\in\mathcal S iff Pf⁡(K)≠0\operatorname{Pf}(K)\ne0. So S=S+⊔S−\mathcal S=\mathcal S_+\sqcup\mathcal S_- with S±={±Pf⁡>0}\mathcal S_\pm=\{\pm\operatorname{Pf}>0\}. Each is open (Pfaffian is continuous) and they are disjoint, so there are at least two components if both are nonempty. They are nonempty: J∈S+J\in\mathcal S_+ and PJPT∈S−PJP^T\in\mathcal S_- for any PP with det⁡P<0\det P<0.

Step 2: normal form. A real skew-symmetric matrix is orthogonally block diagonalisable: K=Q(⨁jsj(01−10))QTK=Q\bigl(\bigoplus_js_j\begin{pmatrix}0&1\\-1&0\end{pmatrix}\bigr)Q^T with Q∈O(n)Q\in O(n) and, if KK is invertible, sj>0s_j>0. Put D=diag⁡(s1,s1,s2,s2,… )D=\operatorname{diag}(\sqrt{s_1},\sqrt{s_1},\sqrt{s_2},\sqrt{s_2},\dots). Then ⨁sj(01−10)=DJDT\bigoplus s_j\begin{pmatrix}0&1\\-1&0\end{pmatrix}=DJD^T, so K=PJPTK=PJP^T with P=QD∈GLn(R)P=QD\in GL_n(\mathbb{R}).

Step 3: each sign set is connected. If K∈S+K\in\mathcal S_+ then K=PJPTK=PJP^T and Pf⁡(K)=det⁡P>0\operatorname{Pf}(K)=\det P>0, so P∈GLn+P\in GL_n^+. Hence S+={PJPT:P∈GLn+}\mathcal S_+=\{PJP^T:P\in GL_n^+\} is the continuous image of the connected group GLn+(R)GL_n^+(\mathbb{R}), so is connected. Similarly S−={PJPT:P∈GLn−}\mathcal S_-=\{PJP^T:P\in GL_n^-\} is the image of the connected set GLn−GL_n^-.

Conclusion. S\mathcal S has exactly the two components S+\mathcal S_+ and S−\mathcal S_-.

Remark. For nn odd every skew matrix is singular (det⁡K=det⁡KT=(−1)ndet⁡K\det K=\det K^T=(-1)^n\det K), so S\mathcal S is empty. For n=2n=2, Skew2≅R\mathrm{Skew}_2\cong\mathbb{R} via K=(0a−a0)K=\begin{pmatrix}0&a\\-a&0\end{pmatrix} and S=R∖{0}\mathcal S=\mathbb{R}\setminus\{0\}, with Pfaffian aa: the two components are a>0a>0 and a<0a<0.

Lesson. The Pfaffian is the "square root of the determinant" that remembers an orientation; it detects the extra component.