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

Consider the space $$HnH_n$$ of hermitian matrices acting on $$Cn\mathbb C^n$$. It contains a subset $$LCnLC_n$$ 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 $$LCnLC_n$$ a submanifold? I suspect that it could, and its codimension is 3. Can we calculate the cohomology ring and some homotopy groups of $$LCnLC_n$$ and $$Hn∖LCnH_n \setminus LC_n$$?

1. $$Hn∖LCnH_n \setminus LC_n$$ is an open subset of $$HnH_n$$, hence submanifold. Moreover it is dense and connected. All these properties are easily seen by considering decomposition $$T=U†DUT=U^{\dagger}DU$$ with $$DD$$ – diagonal and $$UU$$ unitary.
2. Let $$En\mathcal E_n$$ be the space of increasing $$nn$$-tuples of real numbers and let $$Fn\mathcal F_n$$ be the complete flag variety, $$Fn=U(n)U(1)n\mathcal F_n = \frac{U(n)}{U(1)^n}$$. Then $$Hn∖LCnH_n \setminus LC_n$$ is diffeomorphic to the cartesian product $$En×Fn\mathcal E_n \times \mathcal F_n$$. Once again, this is is easy to see using the $$U†DUU^{\dagger}DU$$ decomposition. Elements of $$En\mathcal E_n$$ are the eigenvalues, while elements of $$Fn\mathcal F_n$$ are the eigenspaces.
3. Since $$En≅Rn\mathcal E_n \cong \mathbb R^n$$, we see that $$H∙(Hn∖LCn)≅H∙(Fn)H^{\bullet}(H_n \setminus LC_n) \cong H^{\bullet} (\mathcal F_n)$$. This cohomology group is computed on Wikipedia in the article on generalized flag varieties. I think this description of the topology of $$Hn∖LCnH_n \setminus LC_n$$ is pretty complete and satisfying. However I am also interested in $$LCnLC_n$$ itself.
4. Let $$χT\chi_T$$ be the characteristic polynomial of $$TT$$. Operator $$TT$$ is in $$LCnLC_n$$ if and only if $$χT\chi_T$$ has a double zero. This is equivalent to vanishing of the discriminant $$Δ\Delta$$ of $$χT\chi_T$$, which is easily seen to be a polynomial in the matrix elements of $$TT$$. Thus $$LCnLC_n$$ is an algebraic variety in $$Hn≅Rn2H_n \cong \mathbb R^{n^2}$$.
5. For any $$nn$$ there exist points of $$LCnLC_n$$ on which the first derivative of $$Δ\Delta$$ vanishes. Thus it’s impossible to conclude that $$LCnLC_n$$ is a submanifold using implicit function theorem.
6. My conjecture about codimension $$33$$ is based on the analysis of oribts of $$U(n)U(n)$$ acting on $$HnH_n$$. Namely we need to tune one real number parametrizing $$TT$$ to make it degenerate, but then dimension of the stabilizer of $$TT$$ in $$U(n)U(n)$$ (acting by conjugation) becomes larger at lest by $$22$$. More precisely, if $$TT$$ has $$kk$$ distinct eigenvalues with dimensions of eigenspaces $$g1,...,gkg_1,...,g_k$$, then $$Stab(T)≅U(g1)×...×U(gk)\mathrm{Stab}(T) \cong U(g_1) \times ... \times U(g_k)$$.
7. $$00$$ is in $$LCnLC_n$$ and $$λT∈LCn\lambda T \in LC_n$$ whenever $$T∈LCnT \in LC_n$$ and $$λ∈R\lambda \in \mathbb R$$. In particular $$LCnLC_n$$ is contractible to a point. Clearly the “correct” way of studying the geometry of $$LCnLC_n$$ would be to consider it as a projective variety in $$PRn2−1\mathbb P \mathbb R^{n^2-1}$$. In fact $$Δ\Delta$$ is a homogeneous polynomial of degree $$n(n−1)n(n-1)$$.

Question: “Consider the space $$HnH^n$$ of hermitian matrices acting on $$CnC^n$$. It contains a subset $$LCnLC_n$$ 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 $$LCnLC_n$$ a submanifold?”

Comment: If $$X:=PdkX:=\mathbb{P}^d_k$$ is projective $$dd$$-space over a field and if $$XX$$ parametrize homogeneous polynomials in $$x0,x1x_0,x_1$$ over $$kk$$ of degree $$dd$$, it follows the classical discriminant hypersurface $$Dd(1)D_d(1)$$ parametrizing homogeneous degree $$dd$$ polynomials with “repeated roots” is a singular hypersurface $$Dd(1):=Z(f)⊆XD_d(1):=Z(f)\subseteq X$$. If we write a general homogeneous polynomial of degree $$dd$$ as follows:

$$g:=a0xd0+a1xd−10x1+⋯+adxd1,g:=a_0x_0^d+a_1x_0^{d-1}x_1 + \cdots + a_dx_1^d,$$
it follows $$ff$$ is a homogeneous and irreducible polynomial in the coefficients $$aia_i$$ of $$gg$$. There are “higher discriminants” $$Dd(i)D_d(i)$$ for $$i≥1i \geq 1$$ and a stratification of closed subvarieties

$$Dd(i+1)⊆Dd(i)⊆⋯⊆Dd(1) D_d(i+1) \subseteq D_d(i) \subseteq \cdots \subseteq D_d(1)$$

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

$$π:P1k×⋯×P1k→Pdk\pi: \mathbb{P}^1_k \times \cdots \times \mathbb{P}^1_k \rightarrow \mathbb{P}^d_k$$

identifying $$Pdk≅(P1k)×d/Sd\mathbb{P}^d_k \cong (\mathbb{P}^1_k)^{\times d}/S_d$$ where $$SdS_d$$ is the symmetric group on $$dd$$ letters. The image of the “big diagonal” $$Δ\Delta$$ is the discriminant $$Dd(1)D_d(1)$$. 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

$$π:X:=(A1k)×n→Ank\pi: X:=(\mathbb{A}^1_k)^{\times n} \rightarrow \mathbb{A}^n_k$$

defined by

$$π(a1,..,an):=(s1(ai),..,sn(ai))\pi(a_1,..,a_n):=(s_1(a_i),..,s_n(a_i))$$

where $$sis_i$$ is the $$ii$$‘th symmetric polynomial in the “variables” $$aia_i$$.
If $$ai≠aj∈ka_i \neq a_j\in k$$ for all $$ii$$ and $$y:=π(ai)y:=\pi(a_i)$$ it follows the fiber $$π−1(y)\pi^{-1}(y)$$ has $$n!n!$$ points $$σ(ai),σ∈Sn\sigma(a_i), \sigma \in S_n$$ with $$ai∈ka_i\in k$$.
Hence outside of the big diagonal $$Δ⊆X\Delta \subseteq X$$ it follows $$π\pi$$ has “the same number of elements in the fiber”. If $$x∈Δx\in \Delta$$ and $$y:=π(x)y:=\pi(x)$$ it follows the number of points in $$π−1(y)\pi^{-1}(y)$$ drops. Here you must take care about the notion “point”. If $$kk$$ is not algebraically closed it follows closed points have a residue field that is a finite extension of $$kk$$. Since the map $$π\pi$$ is finite it follows the image of the big diagonal is a closed subvariety of $$Ank\mathbb{A}^n_k$$ and it equals the discriminant.

Example 1: Let $$S:=(A1k)×2S:=(\mathbb{A}^1_k)^{\times 2}$$ and consider the map

$$π:S→A2s1,s2\pi: S\rightarrow \mathbb{A}^2_{s_1,s_2}$$

defined by the map

$$ϕ:k[s1,s2]→k[x,y]\phi: k[s_1,s_2]\rightarrow k[x,y]$$

with $$ϕ(s1):=x+y,ϕ(s2):=xy\phi(s_1):=x+y, \phi(s_2):=xy$$. The discriminant $$D(1)D(1)$$ is defined by $$D(1):=V(s21−4s2)⊆A2s1,s2D(1):=V(s_1^2-4s_2)\subseteq \mathbb{A}^2_{s_1,s_2}$$, and the “inverse image”
is given as follows:

$$π−1(D(1)):=V(s1(x,y)2−4s2(x,y))\pi^{-1}(D(1)):=V(s_1(x,y)^2-4s_2(x,y))$$

and

$$s1(x,y)2−4s2(x,y)=(x−y)2,s_1(x,y)^2-4s_2(x,y)=(x-y)^2,$$

Hence $$Δ:=π−1(D(1)):=Spec(k[x,y]/((x−y)2)\Delta:=\pi^{-1}(D(1)):=Spec(k[x,y]/((x-y)^2)$$. Hence in this case when you calculate the inverse image $$π−1(D(1))\pi^{-1}(D(1))$$ you get a copy of the affine line $$A1k\mathbb{A}^1_k$$ with a non-reduced structure. Hence you must “worry” about the algebraic structure. The $$kk$$-rational points $$Δ(k)\Delta(k)$$ is the following set: a map $$ϕ\phi$$ of $$kk$$-algebras

$$ϕ:k[x,y]/(x−y)2→k\phi: k[x,y]/(x-y)^2 \rightarrow k$$

must fulfill $$ϕ(x):=a,ϕ(y):=b\phi(x):=a, \phi(y):=b$$ and

$$(a−b)2=a2−2ab+b2=0.(a-b)^2=a^2-2ab+b^2=0.$$ Since $$kk$$ is a field, this is the set of $$(a,b)(a,b)$$ with $$a=ba=b$$. Hence $$Δ\Delta$$ is a scheme with the same $$kk$$-rational points as the big diagonal. There is an obvious $$S2S_2$$-action on $$R:=k[x,y]/(x−y)2R:=k[x,y]/(x-y)^2$$ which is trivial on the $$kk$$-rational points of $$RR$$. In this case there is an isomorphism

$$RS2≅k[s1,s2]/(s21−4s2).R^{S_2} \cong k[s_1,s_2]/(s_1^2-4s_2).$$

Hence the discriminant of degree 2 polynomials is a quotient of a non-reduced scheme $$V((x−y)2)V((x-y)^2)$$ by $$S2S_2$$. In general you may realize $$Dd(1)D_d(1)$$ as a quotient of a non-reduced scheme by $$SdS_d$$. 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 $$E:=k{e1,..,en}E:=k\{e_1,..,e_n\}$$ where $$kk$$ is the field of real numbers and let $$KK$$ be the field of complex numbers. Let $$E∗:=k{y1,..,yn}E^*:=k\{y_1,..,y_n\}$$ where $$yi:=e∗iy_i:=e_i^*$$ is the dual basis. It follows

$$Symk(Endk(E)∗)≅A:=k[xij] Sym_k(End_k(E)^*)\cong A:=k[x_{ij}]$$

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

$$GL(E):=Z(H(t))⊆M(n,k):=Spec(A) GL(E):=Z(H(t)) \subseteq M(n,k):=Spec(A)$$

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

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

$$Hrm≅Rn2⊆M(n,C),Hrm\cong \mathbb{R}^{n^2} \subseteq M(n,\mathbb{C}),$$

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 $$χT(t)\chi_T(t)$$ of a matrix $$T∈HrmT\in Hrm$$ you should get real algebraic subvarieties of $$HrmHrm$$. If $$LCn(1)LC_n(1)$$ is the set of matrices in $$HrmHrm$$ with inseparable characteristic polynomial the following holds: Let $$T:=(aij)∈HrmT:=(a_{ij})\in Hrm$$ be a Hermitian matrix and let

$$χT(t):=u0+u1t+⋯+ultl∈R[t]\chi_T(t):=u_0+u_1t+\cdots +u_lt^l \in \mathbb{R}[t]$$

be its characteristic polynomial. It follows $$LCn(1)LC_n(1)$$ is a hypersurface in $$HrmHrm$$ defined as the zero set of a hypersurface $$f(ui)f(u_i)$$ where $$f(ui)f(u_i)$$ is a polynomial in the coeffiencts $$uiu_i$$ of $$χT\chi_T$$. The coefficients $$ui:=ui(aij)u_i:=u_i(a_{ij})$$ are polynomials in the real numbers $$aija_{ij}$$ defining $$TT$$. Hence $$LCn(1)LC_n(1)$$ is the zero set of a polynomial in the “variables” $$aija_{ij}$$, hence $$LCn(1)LC_n(1)$$ is a real algebraic sub variety of $$HrmHrm$$.

Algebraically if $$R:=Z[u0,..,ud]R:=\mathbb{Z}[u_0,..,u_d]$$ is the polynomial ring over the ring of integers in the variables $$uiu_i$$ it follows the discriminant polynomial
$$f(ui)f(u_i)$$ is an irreducible polynomial in $$RR$$. In your case it follows the characteristic polynomial $$χT(t):=P(aij,t)\chi_T(t):=P(a_{ij},t)$$ is a polynomial in the $$aij,ta_{ij},t$$ and substuting in $$f(ui)f(u_i)$$ you get a polynomial

$$F(aij):=f(u0(aij),...,ud(aij))F(a_{ij}):=f(u_0(a_{ij}),..., u_d(a_{ij}))$$

and it may be $$F(aij)F(a_{ij})$$ is an irreducible polynomial in the “variables” $$aija_{ij}$$ in some cases. In general it seems that $$LCn(1):=Z(F(aij))⊆Rn2LC_n(1):=Z(F(a_{ij})) \subseteq \mathbb{R}^{n^2}$$ is a singular algebraic hypersurface that is irreducible in some cases.

Question: “In particular is $$LCnLC_n$$ a submanifold?”

Answer: It seems to me that $$LCn(1)LC_n(1)$$ 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 $$LCn(1):=Z(F)LC_n(1):=Z(F)$$ is a real algebraic variety with a non-empty singular subvariety $$LCn(1)singLC_n(1)_{sing}$$ in general.

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

The open complement $$U:=LCn(1)−LCn(1)singU:=LC_n(1)-LC_n(1)_{sing}$$ has the structure of a real smooth manifold and this manifold may be studied using algebraic and analytic techniques. In general if $$MM$$ is a real algebraic manifold and $$EE$$ is a finite rank algebraic vector bundle on $$MM$$ with an algebraic endomorphism $$ϕ\phi$$, you may construct several types of discriminants $$D(ϕ)D(\phi)$$ of $$ϕ\phi$$ 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).

http://www-users.math.umn.edu/~rober005/publicat/DualVar.p1.html

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.