Topology of “degenerate spectrum submanifold” in the space of hermitian matrices

Consider the space Hn of hermitian matrices acting on Cn. It contains a subset LCn of matrices with degenerate spectrum. I want to know as much as possible about topology and geometry of this set and its complement. In particular is LCn a submanifold? I suspect that it could, and its codimension is 3. Can we calculate the cohomology ring and some homotopy groups of LCn and HnLCn?

  1. HnLCn is an open subset of Hn, hence submanifold. Moreover it is dense and connected. All these properties are easily seen by considering decomposition T=UDU with D – diagonal and U unitary.
  2. Let En be the space of increasing n-tuples of real numbers and let Fn be the complete flag variety, Fn=U(n)U(1)n. Then HnLCn is diffeomorphic to the cartesian product En×Fn. Once again, this is is easy to see using the UDU decomposition. Elements of En are the eigenvalues, while elements of Fn are the eigenspaces.
  3. Since EnRn, we see that H(HnLCn)H(Fn). This cohomology group is computed on Wikipedia in the article on generalized flag varieties. I think this description of the topology of HnLCn is pretty complete and satisfying. However I am also interested in LCn itself.
  4. Let χT be the characteristic polynomial of T. Operator T is in LCn if and only if χT has a double zero. This is equivalent to vanishing of the discriminant Δ of χT, which is easily seen to be a polynomial in the matrix elements of T. Thus LCn is an algebraic variety in HnRn2.
  5. For any n there exist points of LCn on which the first derivative of Δ vanishes. Thus it’s impossible to conclude that LCn is a submanifold using implicit function theorem.
  6. My conjecture about codimension 3 is based on the analysis of oribts of U(n) acting on Hn. Namely we need to tune one real number parametrizing T to make it degenerate, but then dimension of the stabilizer of T in U(n) (acting by conjugation) becomes larger at lest by 2. More precisely, if T has k distinct eigenvalues with dimensions of eigenspaces g1,...,gk, then Stab(T)U(g1)×...×U(gk).
  7. 0 is in LCn and λTLCn whenever TLCn and λR. In particular LCn is contractible to a point. Clearly the “correct” way of studying the geometry of LCn would be to consider it as a projective variety in PRn21. In fact Δ is a homogeneous polynomial of degree n(n1).


Question: “Consider the space Hn of hermitian matrices acting on Cn. It contains a subset LCn of matrices with degenerate spectrum. I want to know as much as possible about topology and geometry of this set and its complement. In particular is LCn a submanifold?”

Comment: If X:=Pdk is projective d-space over a field and if X parametrize homogeneous polynomials in x0,x1 over k of degree d, it follows the classical discriminant hypersurface Dd(1) parametrizing homogeneous degree d polynomials with “repeated roots” is a singular hypersurface Dd(1):=Z(f)X. If we write a general homogeneous polynomial of degree d as follows:

it follows f is a homogeneous and irreducible polynomial in the coefficients ai of g. There are “higher discriminants” Dd(i) for i1 and a stratification of closed subvarieties


and in some cases it follows Dd(i+1)=Dd(i)sing is the subvariety of Dd(i) of singular points. Hence Dd(1) is singular in general. Over a field there is a construction using the symmetric product: There is a canonical map of schemes


identifying Pdk(P1k)×d/Sd where Sd is the symmetric group on d letters. The image of the “big diagonal” Δ is the discriminant Dd(1). This construction gives a geometric construction of the discriminant and various types of stratifications in the algebraic setting. In the affine situation you get a map of algebraic varieties


defined by


where si is the i‘th symmetric polynomial in the “variables” ai.
If aiajk for all i and y:=π(ai) it follows the fiber π1(y) has n! points σ(ai),σSn with aik.
Hence outside of the big diagonal ΔX it follows π has “the same number of elements in the fiber”. If xΔ and y:=π(x) it follows the number of points in π1(y) drops. Here you must take care about the notion “point”. If k is not algebraically closed it follows closed points have a residue field that is a finite extension of k. Since the map π is finite it follows the image of the big diagonal is a closed subvariety of Ank and it equals the discriminant.

Example 1: Let S:=(A1k)×2 and consider the map


defined by the map


with ϕ(s1):=x+y,ϕ(s2):=xy. The discriminant D(1) is defined by D(1):=V(s214s2)A2s1,s2, and the “inverse image”
is given as follows:




Hence Δ:=π1(D(1)):=Spec(k[x,y]/((xy)2). Hence in this case when you calculate the inverse image π1(D(1)) you get a copy of the affine line A1k with a non-reduced structure. Hence you must “worry” about the algebraic structure. The k-rational points Δ(k) is the following set: a map ϕ of k-algebras


must fulfill ϕ(x):=a,ϕ(y):=b and

(ab)2=a22ab+b2=0. Since k is a field, this is the set of (a,b) with a=b. Hence Δ is a scheme with the same k-rational points as the big diagonal. There is an obvious S2-action on R:=k[x,y]/(xy)2 which is trivial on the k-rational points of R. In this case there is an isomorphism


Hence the discriminant of degree 2 polynomials is a quotient of a non-reduced scheme V((xy)2) by S2. In general you may realize Dd(1) as a quotient of a non-reduced scheme by Sd. Hence for the classical discriminant we observe that we need non-reduced schemes in order to realize it as a quotient.

Example 2: In algebra/geometry you are interested a canonical scheme structure on the “discriminant locus” and for this reason you must give a definition that is “intrinsic”. When working with real manifolds you end up with singular spaces.

Let E:=k{e1,..,en} where k is the field of real numbers and let K be the field of complex numbers. Let E:=k{y1,..,yn} where yi:=ei is the dual basis. It follows


and A is naturally the polynomial functions on the vector space Endk(E).
We may realize GL(E):=Spec(B) where B:=A[t]/(tdet(T)1) and


is a smooth hypersurface. The complement D(H(t)):=M(n,k)GL(E) is the “space of singular n×n matrices” and this complement has several types of
stratifications defined in terms ot the characteristic polynomial χT(t) of a matrix T.

You must do something similar for Hermitian matrices, that is: Instead of the group GL(E) you must let HrmM(n,K) be the set of n×n Hermitian matrices and consider a stratification on Hrm. It seems there is an isomorphism


hence the hermitian matrices form an affine real space which is an algebraic variety. When you put conditions on the multiplicity of zeros of the characteristis polynomial χT(t) of a matrix THrm you should get real algebraic subvarieties of Hrm. If LCn(1) is the set of matrices in Hrm with inseparable characteristic polynomial the following holds: Let T:=(aij)Hrm be a Hermitian matrix and let


be its characteristic polynomial. It follows LCn(1) is a hypersurface in Hrm defined as the zero set of a hypersurface f(ui) where f(ui) is a polynomial in the coeffiencts ui of χT. The coefficients ui:=ui(aij) are polynomials in the real numbers aij defining T. Hence LCn(1) is the zero set of a polynomial in the “variables” aij, hence LCn(1) is a real algebraic sub variety of Hrm.

Algebraically if R:=Z[u0,..,ud] is the polynomial ring over the ring of integers in the variables ui it follows the discriminant polynomial
f(ui) is an irreducible polynomial in R. In your case it follows the characteristic polynomial χT(t):=P(aij,t) is a polynomial in the aij,t and substuting in f(ui) you get a polynomial


and it may be F(aij) is an irreducible polynomial in the “variables” aij in some cases. In general it seems that LCn(1):=Z(F(aij))Rn2 is a singular algebraic hypersurface that is irreducible in some cases.

Question: “In particular is LCn a submanifold?”

Answer: It seems to me that LCn(1) is singular in general, hence it is not a “manifold”/smooth variety. Discriminants appear in algebraic geometry, differential geomeary and complex analysis and they are singular spaces in general.

In your case the discriminant LCn(1):=Z(F) is a real algebraic variety with a non-empty singular subvariety LCn(1)sing in general.

The polynomial F defining LCn(1) gives rise to the Jacobi ideal
J(F) which is the ideal generated by F and all its partial derivatives. If A:=R[aij]/(F) is the coordinate ring of LCn(1) it follows the subvariety of singular points S:=LCn(1)sing is defined by the ideal J(F). To prove that S is nonempty you must verify that a polynomial with a root of multiplicity 3 defines a point in S.

The open complement U:=LCn(1)LCn(1)sing has the structure of a real smooth manifold and this manifold may be studied using algebraic and analytic techniques. In general if M is a real algebraic manifold and E is a finite rank algebraic vector bundle on M with an algebraic endomorphism ϕ, you may construct several types of discriminants D(ϕ) of ϕ and these discriminants are singular algebraic varieties in general. These types of discriminants are much studied in algebra and geometry:

References: Busé, Laurent; Jouanolou, Jean-Pierre
On the discriminant scheme of homogeneous polynomials. Zbl 1302.13028
Math. Comput. Sci. 8, No. 2, 175-234 (2014).

There is another recent paper (the paper may be found online) which may give references to where it is proved that the classical discriminant of degree d polynomials is singular:

Demazure, Michel, Resultant, discriminant. (Résultant, discriminant.) Enseign. Math. (2) 58, No. 3-4, 333-373 (2012).

I believe there is an algebraic proof of this and you may find interesting references in the above papers and links. On page 6 in the above linked paper you will find a discussion of the classical discriminant that may be interesting. Another reference is the following book:

Gelfand, I. M.; Kapranov, M. M.; Zelevinsky, A. V.
Discriminants, resultants, and multidimensional determinants.
Modern Birkhäuser Classics. Boston, MA: Birkhäuser

It contains a discussion of the classical discriminant and many references.

Source : Link , Question Author : Blazej , Answer Author : hm2020

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