Let n be even and let S={K∈Skewn:detK=0}. Show that S has exactly two connected components, distinguished by the sign of the Pfaffian Pf(K).
Hint
Every nondegenerate skew matrix has the form K=PJPT with P∈GLn(R) and J the standard block matrix, and Pf(PJPT)=detP⋅Pf(J).
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Solution
Facts used. The Pfaffian is a polynomial in the entries of K with Pf(K)2=detK and Pf(PKPT)=det(P)Pf(K). For J=⨁j=1n/2(0−110) one has Pf(J)=1. We also use that GLn+(R) and GLn−(R) are connected (standard).
Step 1: the Pfaffian separates. Since detK=Pf(K)2, K∈S iff Pf(K)=0. So S=S+⊔S− with S±={±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+ and PJPT∈S− for any P with detP<0.
Step 2: normal form. A real skew-symmetric matrix is orthogonally block diagonalisable: K=Q(⨁jsj(0−110))QT with Q∈O(n) and, if K is invertible, sj>0. Put D=diag(s1,s1,s2,s2,…). Then ⨁sj(0−110)=DJDT, so K=PJPT with P=QD∈GLn(R).
Step 3: each sign set is connected. If K∈S+ then K=PJPT and Pf(K)=detP>0, so P∈GLn+. Hence S+={PJPT:P∈GLn+} is the continuous image of the connected group GLn+(R), so is connected. Similarly S−={PJPT:P∈GLn−} is the image of the connected set GLn−.
Conclusion.S has exactly the two components S+ and S−.
Remark. For n odd every skew matrix is singular (detK=detKT=(−1)ndetK), so S is empty. For n=2, Skew2≅R via K=(0−aa0) and S=R∖{0}, with Pfaffian a: the two components are a>0 and a<0.
Lesson. The Pfaffian is the "square root of the determinant" that remembers an orientation; it detects the extra component.