HERMITIAN DISTANCE DEGREE OF
UNITARY-INVARIANT MATRIX VARIETIES

Nikhil Ken

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Nikhil Ken. HERMITIAN DISTANCE DEGREE OF UNITARY-INVARIANT MATRIX VARIETIES. 2026. hal-05536805

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# HERMITIAN DISTANCE DEGREE OF UNITARY-INVARIANT MATRIX VARIETIES

NIKHIL KEN

**ABSTRACT.** We study the Hermitian distance degree, a real enumerative invariant counting critical points of the squared Hermitian distance function, for matrix varieties invariant under left and right unitary actions. For such a variety $M \subset \mathbb{C}^{n \times t}$, we prove that its Hermitian distance degree equals the real Euclidean distance degree of the associated absolutely symmetric variety of singular values. Equivalently, for a generic data matrix, Hermitian distance critical points on $M$ are obtained by lifting Euclidean distance critical points from the singular-value slice. We also establish a Hermitian slicing theorem, paralleling the Bik-Draisma principle, which reduces the critical point count to a diagonal slice. As a motivating example, we recover a geometric Hermitian analogue of the Eckart-Young theorem.

## 1. INTRODUCTION

In this article, we study a real enumerative invariant attached to Hermitian distance critical points, which we call the Hermitian distance degree (introduced and studied in detail in [Fur25]), focusing on matrix varieties invariant under left and right unitary actions. Recall that the ED degree of an algebraic variety $X \subset \mathbb{C}^n$ is the number of critical points in $X^{\text{reg}}$ of the squared distance function

$$d_u(x) = \|x - u\|^2,$$

arising from a Euclidean inner product and a general data point $u \in \mathbb{C}^n$; see [DHO$^{+16}$], [DLOT17], [HW19], [Sod21] and [JB15].

In the Hermitian setting, for matrices $A, B \in \mathbb{C}^{n \times t}$, we consider the Hermitian inner product

$$\langle A, B \rangle = \text{Tr}(AB^*),$$

where $B^*$ is the conjugate transpose of $B$. The Hermitian distance degree is a real enumerative invariant attached to the Hermitian distance critical points. Concretely, we view $M \subseteq \mathbb{C}^{n \times t} = V$ as a real algebraic set in $V^\mathbb{R} \cong \mathbb{R}^{2nt}$ and use the real inner product $q = \text{Re}\langle \cdot, \cdot \rangle$. Since we work in $V^\mathbb{R}$ with $q = \text{Re}\langle \cdot, \cdot \rangle$, we count real critical points; complex critical points of the complexified equations are not critical points of the real problem.

For a data matrix $Y \in \mathbb{C}^{n \times t}$, let $d_Y(X) = \|X - Y\|^2$ be the squared distance function with respect to $q$. For $Y$ outside a certain semialgebraic discriminant locus, the number of real critical points of $d_Y$ on $M^{\text{reg}}$ is finite and constant on the chamber of $Y$; we denote this chamber-wise constant by $\text{HDdeg}(M; Y)$. The set of all such chamberwise values is denoted by

$$\text{HDdeg}(M) := \{\text{HDdeg}(M; Y) : Y \in V^\mathbb{R} \setminus \Delta_M\} \subset \mathbb{N}.$$

Let $f: \mathbb{C}^{n \times t} \to \mathbb{R}$ be defined by

$$f(x) = \|x - y\|^2 = \langle x - y, x - y \rangle,$$

where $\langle \cdot, \cdot \rangle$ is the Hermitian inner product. For $f(x) = \|x - y\|^2$ induced by the Hermitian product, the differential satisfies

$$df_x(v) = \langle v, x - y \rangle + \langle x - y, v \rangle = 2 \text{Re}\langle x - y, v \rangle.$$

This observation means that the Hermitian distance problem is governed by the real symmetric bilinear form

$$(\cdot, \cdot)_\mathbb{R} := \text{Re}\langle \cdot, \cdot \rangle$$

on the real vector space $V^\mathbb{R}$. In particular, Hermitian distance critical points are precisely Euclidean distance critical points for $X$, viewed as a real algebraic set in $V^\mathbb{R}$ with respect to $(\cdot, \cdot)_\mathbb{R}$, and we count only the real critical points of this real optimization problem; hence the relevant enumerative invariant is a real Euclidean distance degree, denoted $\text{REDdeg}$.
In [DLOT17], the authors study the ED degree of matrix varieties invariant under the action of orthogonal groups, and show that it coincides with the ED degree of a variety determined by the singular values. This idea was further generalized in [AJ18], where it was shown that the ED degree of a variety X with a suitable group action can sometimes be computed by slicing X with an appropriate subspace.

Building on these ideas, we show in Theorem 23 that the HD degree of a Unitary-invariant matrix variety $M \subset \mathbb{C}^{n \times t}$ equals the real ED degree of the associated absolutely symmetric variety $S \subset \mathbb{R}^n$ of singular values. Concretely, for a generic data matrix $Y = U \text{diag}(y)V^*$, one has

$$ \text{HDdeg}(M; Y) = \text{REDdeg}(S; y). $$

Moreover, this common value is constant on chambers of $\mathbb{R}^n$; when the chamber is clear from context, we denote it simply by $\text{HDdeg}(M)$ (or equivalently $\text{REDdeg}(S)$).

Beyond the numerical equality, Theorem 23 gives a concrete description of the critical points themselves. For a generic data matrix $Y = U \text{diag}(y)V^*$, every Hermitian distance critical point on $M$ is obtained by lifting a real Euclidean distance critical point of $y$ on $S$, using the same singular vectors $U, V$. Since the counting problem is genuinely real, the number of critical points can depend on the chamber of $y$, and may jump when $y$ crosses the ED discriminant. We illustrate this behaviour explicitly in Example 24, 25.

We also prove a Hermitian analogue of the slicing principle of Bik-Draisma by passing to $V^\mathbb{R}$ and exploiting that unitary representations are orthogonal with respect to $q = \text{Re}\langle \cdot, \cdot \rangle$.

As a motivating case, we revisit the Eckart-Young theorem (Theorem 5) geometrically for Hermitian distance, recovering the critical points of the Hermitian distance function to the rank-k determinantal variety (compare [OP15] for the Euclidean setting). This becomes a special case of Theorem 23. Hermitian distance critical equations also arise in the study of closest product states and geometric measures of entanglement; see [HS10].

**Conventions.** Throughout we identify the complex vector space $V = \mathbb{C}^{n \times t}$ as a real vector space

$$ V^{\mathbb{R}} \cong \mathbb{R}^{2nt}, $$

and we equip $V^{\mathbb{R}}$ with the real Euclidean inner product

$$ q(A, B) := \text{Re}\langle A, B \rangle = \text{Re}(\text{Tr}(AB^*)). $$

All algebraic subsets of $V$ will be viewed as real algebraic sets in $V^{\mathbb{R}}$ when counting (real) distance critical points. For $n \le t$, we write $\text{diag}(d)$ for the rectangular diagonal matrix in $\mathbb{C}^{n \times t}$ with $(\text{diag}(d))_{ii} = d_i$ for $1 \le i \le n$ and all other entries zero.

**Ambient-space subscripts.** When we write $\text{REDdeg}$ or $\text{HDdeg}$ with a subscript, the subscript indicates the ambient real inner-product space in which the critical points are computed. When the ambient space is clear from context, we omit the subscript.

## ACKNOWLEDGEMENTS

This work was supported by the European Union's Horizon Europe programme under the Marie Skłodowska-Curie Actions, HORIZON-MSCA-2023-DN-JD (grant agreement No. 101120296, TENORS). I thank Giorgio Maria Ottaviani for comments on multiple drafts, and Davide Furchì for helpful discussions on the Hermitian distance degree.

## 2. PRELIMINARIES

**Theorem 1.** [GVL13](2.4.4) Let $A$ be a matrix in $\mathbb{C}^{m \times n}$. Then $A$ can be factored as

$$ A = U \Sigma V^* $$

where $U \in \mathbb{C}^{m \times m}$ is unitary, $V \in \mathbb{C}^{n \times n}$ is unitary, and $\Sigma \in \mathbb{R}^{m \times n}$ has the form

$$ \Sigma = \text{diag}(s_1, s_2, \dots, s_p), $$

where $p = \min(m, n)$. The $s_i$s are called the singular values of $A$, and they are the square roots of the eigenvalues of $AA^*$
**Definition 2.** Let $S \subseteq \mathbb{R}^n$ be a real algebraic set and let $y \in \mathbb{R}^n$. A smooth point $x \in S^{\text{reg}}$ is an ED critical point of $y$ on $S$ if

$$x - y \perp T_x S, \quad \text{i.e.} \quad (x - y)^{\top} a = 0 \text{ for all } a \in T_x S.$$

Analogously, let $M \subseteq \mathbb{C}^{n \times t}$ be a matrix variety and $Y \in \mathbb{C}^{n \times t}$. A smooth point $X \in M^{\text{reg}}$ is an HD critical point of $Y$ on $M$ if

$$\operatorname{Re} \operatorname{Tr}((Y - X)A^*) = 0 \quad \text{for all } A \in T_X M.$$

**Lemma 3.** If $A_1 = u_1 v_1^*$ and $A_2 = u_2 v_2^*$ are rank-one matrices, then the Hermitian inner product given by $\langle A_1, A_2 \rangle = \operatorname{Tr}(A_1 A_2^*)$ is also given by

$$\langle A_1, A_2 \rangle = \langle u_1, u_2 \rangle \langle v_1, v_2 \rangle.$$

*Proof.*

$$\operatorname{Tr} \left[ \begin{pmatrix} u_{11} \\ \vdots \\ u_{m1} \end{pmatrix} (v_{11} \quad \cdots \quad v_{1n}) \cdot \begin{pmatrix} \overline{v_{21}} \\ \vdots \\ \overline{v_{2n}} \end{pmatrix} (\overline{u_{21}} \quad \cdots \quad \overline{u_{2m}}) \right]$$

$$\text{So we have } \langle A_1, A_2 \rangle = \sum_i u_{1i} (\sum_k v_{1k} \overline{v_{2k}}) \overline{u_{2i}} = \sum_i u_{1i} \overline{u_{2i}} \sum_k v_{1k} \overline{v_{2k}} = \langle u_1, u_2 \rangle \langle v_1, v_2 \rangle \quad \square$$

**Lemma 4.** Let $B \in \mathbb{C}^{m \times n}$. If $\langle B, \mathbb{C}^m \otimes v \rangle = 0$, then $\langle \operatorname{Row}(B), v \rangle = 0$. Similarly
If $\langle B, u \otimes \mathbb{C}^n \rangle = 0$, then $\langle \operatorname{Col}(B), u \rangle = 0$.

*Proof.* Let $e_1, \dots, e_m$ be the canonical basis for $\mathbb{C}^m$ then $\langle B, e_i \otimes v \rangle = 0 \quad \forall i = 1, \dots, m$. Then we have $\operatorname{Tr}(B(\bar{v} \otimes e_i)) = 0 = \operatorname{Tr}(B(0, \dots, \bar{v}, \dots, 0)) = \langle B^i, v \rangle$ where $B^i$ is the i-th row of B. By a similar argument, if $\langle B, u \otimes \mathbb{C}^n \rangle = 0$, then $\langle \operatorname{Col}(B), u \rangle = 0$. $\square$

**Theorem 5 (Eckart-Young for HD).** Let $A \in \mathbb{C}^{m \times n}$ be a matrix of rank $r$, and let its singular value decomposition be

$$A = U \Sigma V^*, \quad \Sigma = \operatorname{diag}(s_1, \dots, s_r, 0, \dots, 0),$$

with singular values $s_1 \ge \dots \ge s_r > 0$.
For $1 \le i \le r$, define

$$\Sigma_i = \operatorname{diag}(0, \dots, 0, s_i, 0, \dots, 0),$$

so that

$$A = U \Sigma_1 V^* + U \Sigma_2 V^* + \dots + U \Sigma_r V^*.$$

Then the critical points of the Hermitian distance function from A to the variety of matrices of rank at most $k$ are precisely the matrices of the form

$$U (\Sigma_{i_1} + \dots + \Sigma_{i_k}) V^*, \quad 1 \le i_1 < \dots < i_k \le r.$$

In particular, if the nonzero singular values of A are distinct, then for each $k \le r$ there are exactly $\binom{r}{k}$ such critical points.

Before proving this theorem, we will talk about how the secant varieties of rank one matrices are defined which gives a geometric interpretation for the rank of matrices and how their tangent spaces behave.

**Definition 6.** Let $X \subset \mathbb{P}V$ be an irreducible variety. The $k$-secant variety of $X$ is defined by:

$$\sigma_k(X) := \overline{\bigcup_{p_1, \dots, p_k \in X} \mathbb{P}\operatorname{Span}\{p_1, \dots, p_k\}}$$

where $\mathbb{P}\operatorname{Span}\{p_1, \dots, p_k\}$ is the smallest projective linear space containing $p_1, \dots, p_k$, and the overbar denotes the Zariski closure.

If we consider $M_1$ to be the rank one matrices over a field $K$ or decomposable tensors in $K^m \otimes K^n$ then $\sigma_r(M_1)$ is just the variety of matrices with rank less than or equal to $r$. This is due to the fact any rank $r$ matrix can be written as sum of rank one matrices. The tangent spaces to the secant varieties can be understood with the help of the following lemma due to Terracini.
**Lemma 7 (Terracini).** Let $z \in \mathbb{P}Span\{p_1, \dots, p_k\}$ where $p_1, \dots, p_k$ be general points. Then:

$$T_z\sigma_k(X) = \mathbb{P}Span\{T_{p_1}X, \dots, T_{p_k}X\}$$

This lemma helps us understand the tangent spaces of $M_r$ (the variety of matrices with rank at most r) with the help of the singular value decomposition. Consider a curve $\gamma(t) = v(t) \otimes u(t)$ with $\gamma(0) = v \otimes u$. The derivative at t=0 is given by $u'(0) \otimes v + u \otimes v'(0)$ where $u'(0)$ and $v'(0)$ are arbitrary vectors so the tangent space is given by $\mathbb{C}^m \otimes u + v \otimes \mathbb{C}^n$ at $u \otimes v$. Now the tangent space to $M_r$ at a point $U(\Sigma_1 + \dots + \Sigma_r)V^* = \sum_{i=1}^r s_i(u_i \otimes v_i^*)$. (where $u_i, v_i$ are column vectors of U and V) can be described, by the Terracini Lemma, as

$$T_{U(\Sigma_1+\dots+\Sigma_r)V^*} M_r = T_{U\Sigma_1V^*} M_1 + \dots + T_{U\Sigma_rV^*} M_1 = T_{s_1u_1\otimes v_1^*} M_1 + \dots + T_{s_ru_r\otimes v_r^*} M_1 = (\mathbb{C}^m \otimes v_1^* + u_1 \otimes \mathbb{C}^n) + \dots + (\mathbb{C}^m \otimes v_r^* + u_r \otimes \mathbb{C}^n). \text{ Now we can show the proof of theorem 5.}$$

**Proof of Theorem 5.** The matrix $U(\Sigma_{i_1} + \dots + \Sigma_{i_k})V^*$ is a critical point of the distance function from A to the variety $M_k$ if and only if the vector

$$A - U(\Sigma_{i_1} + \dots + \Sigma_{i_k})V^*$$

is orthogonal to the tangent space.

$$T_{U(\Sigma_{i_1}+\dots+\Sigma_{i_k})V^*} M_k = (\mathbb{C}^m \otimes v_{i_1}^* + u_{i_1} \otimes \mathbb{C}^n) + \dots + (\mathbb{C}^m \otimes v_{i_k}^* + u_{i_k} \otimes \mathbb{C}^n).$$

From the SVD of A, we have

$$A - (U(\Sigma_{i_1} + \dots + \Sigma_{i_k})V^*) = U(\Sigma_{j_1} + \dots + \Sigma_{j_l})V^* = s_{j_1}u_{j_1} \otimes v_{j_1}^* + \dots + s_{j_l}u_{j_l} \otimes v_{j_l}^*$$

where $\{j_1, \dots, j_l\}$ is the set of indices given by the difference $\{1, \dots, r\} \setminus \{i_1, \dots, i_k\}$. Let $e_1, \dots, e_m$ be the canonical basis of $\mathbb{C}^m$, then by Lemma 3 We have

$$\langle s_{j_h} u_{j_h} \otimes v_{j_h}^*, e_l \otimes v_{i_s}^* \rangle = s_{j_h} \langle u_{j_h}, e_l \rangle \langle v_{j_h}, v_{i_s} \rangle = 0$$

since $v_{j_h}$ and $v_{i_s}$ are distinct columns of the unitary matrix $V$. Thus, the matrices $U\Sigma_{j_h}V^*$ are orthogonal to the spaces $\mathbb{C}^m \otimes V_{i_s}^*$.

In a similar way, since $U$ is a unitary matrix, the matrices $U\Sigma_{j_h}V^*$ are orthogonal to the spaces $u_{i_s} \otimes \mathbb{C}^n$. Therefore, $A - (U(\Sigma_{i_1} + \dots + \Sigma_{i_k})V^*)$ is orthogonal to the tangent space, and $U(\Sigma_{i_1} + \dots + \Sigma_{i_k})V^*$ is a critical point. Now we will show all critical points are of this form, Let $B \in M_k$ be such that $A - B$ is orthogonal to the tangent space $T_B(M_k)$. We consider the SVD of B and A-B, let $B = U'(\Sigma'_1 + \dots + \Sigma'_k)V'^*$, and $A - B = U''(\Sigma''_1 + \dots + \Sigma''_k)V''^*$ with $\Sigma'_k \neq 0$ and $\Sigma''_k \neq 0$. Since A-B is orthogonal to $T_B(M_k)$ we get that $\langle \mathrm{Col}(A - B), u'_s \rangle = 0 = \langle \mathrm{Row}(A - B), v'_s \rangle$ for $s = 1, \dots, k$. In particular, $\mathrm{Col}(A - B)$ is a vector subspace of $\mathrm{Span}\{u'_1, \dots, u'_k\}^\perp$ and has dimension at most $m - k$, while $\mathrm{Row}(A - B)$ is a vector subspace of $\mathrm{Span}\{\bar{v}'_1, \dots, \bar{v}'_k\}^\perp$ and has dimension at most $n - k$, so that $l \leq \min\{m, n\} - k$. From the equality

$$A - B = (u''_1, \dots, u''_l, 0, \dots, 0)(\Sigma''_1 + \dots + \Sigma''_l)V''^*$$

we get

$$\mathrm{Col}(A - B) \subset \mathrm{Span}\{u''_1, \dots, u''_l\}$$

and equality holds because the rank of A-B is 1. In a similar way,

$$\mathrm{Row}(A - B) = \mathrm{Span}\{\bar{v}''_1, \dots, \bar{v}''_l\}.$$

This implies that the orthonormal columns $u''_1, \dots, u''_l, u'_1, \dots, u'_k$ can be completed with orthonormal $m - l - k$ columns of $\mathbb{C}^m$ to obtain a unitary $m \times m$ matrix $U$, while the orthonormal columns $v''_1, \dots, v''_l, v'_1, \dots, v'_k$ can be completed with orthonormal $n - l - k$ columns of $\mathbb{C}^n$ to obtain a unitary $n \times n$ matrix $V$. We get

$$A - B = U \begin{pmatrix} \Sigma'' & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} V^*,$$
$$B = U \begin{pmatrix} 0 & 0 & 0 \\ 0 & \Sigma' & 0 \\ 0 & 0 & 0 \end{pmatrix} V^*,$$

where

$$\Sigma'' = \operatorname{diag}(s_1'', \dots, s_l''), \quad \Sigma' = \operatorname{diag}(s_1', \dots, s_k').$$

Thus $B$ is obtained by keeping exactly $k$ of the rank one summands $U\Sigma_iV^*$ in the SVD expansion of $A$, hence it has the required form. □

**Lemma 8.** *Assume we have a complex subspace $W$ of a vector space $V$ and a Hermitian inner product. If $v \in V$ and $\operatorname{Re} \langle v, w \rangle = 0 \ \forall w \in W$ then $\langle v, w \rangle = 0 \ \forall w \in W$*

*Proof.* Since $W$ is closed under multiplication by $i$, Consider $\operatorname{Re} \langle v, iw \rangle = -i\operatorname{Im} \langle v, w \rangle = 0$. Hence $\langle v, w \rangle = 0$ □

**Remark 9.** If $T_X M$ is a complex linear subspace of $V$, then the condition $\operatorname{Re}\langle Y - X, A \rangle = 0$ for all $A \in T_X M$ implies $\langle Y - X, A \rangle = 0$ for all $A \in T_X M$ (Lemma 8). If $T_X M$ is only real linear, this implication need not hold (see Remark 27).

### 3. MATRICES INVARIANT UNDER UNITARY ACTIONS AND CRITICAL POINTS

Let $M \subset \mathbb{C}^{n \times t}$ be an algebraic variety of matrices invariant under the left and right actions of unitary matrices, i.e.

$$UMV^* = M \quad \text{for all } U \in U(n), V \in U(t),$$

and, without loss of generality, assume $n \le t$.

**Definition 10.** Let $\Pi_n^{\pm}$ be the group of signed permutations, with cardinality $2^n n!$. A set $S \subseteq \mathbb{R}^n$ is absolutely symmetric if

$$S = \pi S \quad \text{for all } \pi \in \Pi_n^{\pm}.$$

For any set $S \subseteq \mathbb{R}^n$, its absolute symmetrization is

$$\Pi_n^{\pm} S := \{\pi x : \pi \in \Pi_n^{\pm}, x \in S\}.$$

If $M$ is a Unitary-invariant matrix variety, define

$$\sigma(M) := \{x \in \mathbb{R}^n : \operatorname{diag}(x) \in M\}.$$

If $S$ is absolutely symmetric, define

$$\sigma^{-1}(S) := \{U \operatorname{diag}(x) V^* : x \in S, U \in U(n), V \in U(t)\}.$$

**Definition 11** (Real ED degree). Let $S \subset \mathbb{R}^n$ be a real algebraic set. For a generic point $y \in \mathbb{R}^n$, consider the squared distance function $d_y(x) = \|x - y\|^2$ restricted to $S$. The number of real critical points of $d_y$ is finite and constant on each chamber of $y$ (i.e. on each connected component of the complement of the ED-discriminant, see [DHO$^{+16}$]). We denote this chamber-wise constant in the chamber of $y$ by $\operatorname{REDdeg}(S; y)$ and the set of all such possible values by $\operatorname{REDdeg}(S) \subset \mathbb{N}$.

**Theorem 12.** A set $M \subset \mathbb{C}^{n \times t}$ is unitary invariant if and only if there exists an absolutely symmetric set $S \subset \mathbb{R}^n$ such that

$$S = \sigma(M) \quad \text{and} \quad M = \sigma^{-1}(S).$$

*Proof.* Assume $M$ is unitary invariant. Let $A \in M$ and write an SVD

$$A = U \Sigma V^*, \quad \Sigma = \operatorname{diag}(\sigma_1, \dots, \sigma_n),$$

(with $n \le t$ so $\Sigma \in \mathbb{C}^{n \times t}$ in the usual rectangular-diagonal sense). Since $M$ is invariant, $\Sigma = U^*AV \in M$, hence $(\sigma_1, \dots, \sigma_n) \in \sigma(M) = S$. Moreover, for any signed permutation $\pi \in \Pi_n^{\pm}$, we can realize $\pi$ by multiplying $\Sigma$ on the left and right by suitable unitary signed permutation matrices; therefore $\operatorname{diag}(\pi\sigma) \in M$ and hence $\pi\sigma \in S$. Thus $S$ is absolutely symmetric.

Conversely, assume $S$ is absolutely symmetric and set $M := \sigma^{-1}(S)$. If $A \in M$, then $A = U\operatorname{diag}(x)V^*$ for some $x \in S$ and unitary $U, V$. For any $U' \in U(n)$ and $V' \in U(t)$ we have

$$U'AV'^* = (U'U)\operatorname{diag}(x)(V'V)^* \in \sigma^{-1}(S) = M,$$

since products of unitary matrices are unitary. Hence $M$ is unitary invariant. □
*Example 13 (Unitary Group).* Consider first the set

$$M = U(n) \subset \mathbb{C}^{n \times n}.$$

It is clearly invariant under left and right multiplication by unitary matrices. Since every unitary matrix has all singular values equal to 1, the corresponding absolutely symmetric set is

$$S = \{x \in \mathbb{R}^n : x_i = 1 \text{ for all } i\}.$$

For any $A \in U(n)$, an SVD has the form $A = U\Sigma V^*$ with $\Sigma = I_n$. Because $\Sigma \in \sigma^{-1}(S)$ and products of unitary matrices are Unitary, we obtain $\sigma^{-1}(S) = U(n)$, as expected.

*Example 14 (Rank Variety).* Consider the determinantal variety

$$M_r = \{A \in \mathbb{C}^{n \times t} : \mathrm{rank}(A) \le r\}.$$

This set is also invariant under left and right unitary actions. A matrix has rank at most $r$ precisely when its singular-value vector has at most $r$ nonzero entries, so the corresponding absolutely symmetric set is

$$S = \{x \in \mathbb{R}^n : \#\{i : x_i \neq 0\} \le r\}.$$

Hence $\sigma^{-1}(S)$ consists exactly of the matrices whose singular values lie in this set, and therefore $\sigma^{-1}(S) = M_r$.

**Lemma 15.** The tangent space at a point $M \in U(n)$ is given by $T_M(U(n)) = \{ZM : Z = -Z^*\} = \{MZ : Z = -Z^*\}$

*Proof.* Consider $F: \mathbb{C}^{n \times n} \to \mathbb{C}^{n \times n}$ given by $M \to MM^*$. Consider $B \in \mathbb{C}^{n \times n}$ and $t \in \mathbb{R}$.

$$(M + tB)(M + tB)^* = MM^* + t(BM^* + MB^*) + t^2BB^*$$

Therefore $T_M(U(n))$ is given by the kernel of the map $B \to MB^* + BM^*$. Consider $Z=MB^*$, then $Z = -Z^*$. If we consider the map $M \to M^*M$ we get the second representation of the tangent space as well.

**Lemma 16.** Consider a matrix $A \in \mathbb{C}^{n \times t}$ and let $D=\mathrm{diag}(d) \in \mathbb{C}^{n \times t}$ with $d_i \neq 0$ and $|d_i|^2$ pairwise distinct. If $AD^*$ and $D^*A$ are Hermitian then $A$ is a diagonal matrix. In particular if $D$ is a real diagonal matrix then $A$ is a real diagonal matrix

*Proof.* $AD^*$ being Hermitian gives us that $\overline{A_{ij}d_j} = A_{ji}\overline{d_i}$ and $D^*A$ being Hermitian gives us that $\overline{A_{ij}d_i} = A_{ji}\overline{d_j}$ for i,j=1 to n and $A_{ij}\overline{d_j} = 0$ for i=1 to n and j>n. This implication in particular gives if entries of D is real so is the entries of A. Therefore we have

$$\overline{A_{ij}d_i d_j} = A_{ji} \overline{d_i d_i}$$

$$\overline{A_{ij}d_i d_j} = A_{ji} \overline{d_j d_j}$$

Comparing these identities gives $A_{ij}(|d_i|^2 - |d_j|^2) = 0$ for all $i, j$. But since $d_i \overline{d_i}$ are distinct for each i we have that $A_{ij} = 0$ for $i \neq j$ and since $d_i \neq 0$ we have $A_{ij} = 0$ for $i = 1$ to n and $j>n$. Thus we have that A is diagonal.

**Theorem 17.** Fix an $U(n) \times U(t)$-invariant matrix variety $M \subseteq \mathbb{C}^{n \times t}$. Consider a matrix $Y \in \mathbb{C}^{n \times t}$ such that the eigenvalues of $YY^*$ are nonzero and distinct. Let $Y=UDV^*$ be an SVD of $Y$ with $U, V$ unitary, then any HD critical point $X$ of $Y$ with respect to $M$ admits a decomposition $X= UAV^*$ where $A$ is a real diagonal matrix.

*Proof.* Consider a singular value decomposition of $Y = UDV^*$ for $U \in U(n)$, $V \in U(t)$ and $D \in \mathbb{C}^{n \times t}$. Let $X$ be an HD critical point of $Y$ with respect to $M$. Let $A=U^*XV$, consider the map from $F: U(n) \to M$ given by $W \to WAV^*$

$$[\nabla F(U)](B) = BAV^* \in T_X(M),$$

for any $B \in T_U(U(n))$. By Lemma 15, we may write $B = UZ$ for a skew-Hermitian matrix $Z$,

$$UZAV^* \in T_X(M).$$

we see that the tangent space of $M$ at $X$ contains

$$\{UZAV^* : Z^* = -Z\}.$$
Then, by the definition of an HD critical point, we have

$$ \mathrm{ReTr}((Y - X)(UZAV^*)^*) = 0 $$

for any skew-Hermitian matrix Z.

$$ 0 = \mathrm{ReTr} (U(D - A)V^*VA^*Z^*U^*) = \mathrm{ReTr} ((D - A)A^*Z^*). $$

by Lemma 22 this means $(D - A)A^*$ is Hermitian also note $(D - A)A^*$ is 0 when the imaginary part of the trace vanishes as well. Since $AA^*$ is Hermitian, we have that $DA^*$ is Hermitian; therefore the transpose $AD^*$ is Hermitian. Similarly we will show $D^*A$ is Hermitian By considering $F:U(t) \to M$ given by $W \mapsto UAW^*$, we get, as above, that

$$ \{UAZ^*V^*: Z^* = -Z\} \subseteq T_X(M). $$

It follows that

$$ 0 = \mathrm{ReTr}((U(D - A)V^*)^*UAZ^*V^*) = \mathrm{ReTr}((D - A)^*AZ^*) $$

for any skew-Hermitian matrix Z, $(D - A)^*A$ is Hermitian. Again, since $A^*A$ is Hermitian, we get that $D^*A$ is Hermitian. Since $AD^*$ and $D^*A$ are both Hermitian, we conclude that A is real diagonal by Lemma 16. □

**Lemma 18.** Let $S \subseteq \mathbb{R}^n$ be a $\Pi_n^\pm$-invariant real algebraic set and set

$$ M := \sigma^{-1}(S) = \{ U \mathrm{diag}(x)V^*: x \in S, U \in U(n), V \in U(t) \}. $$

If $Q \subseteq S$ is Euclidean dense in $S$, then

$$ N_Q := \{ U \mathrm{diag}(x)V^*: x \in Q, U \in U(n), V \in U(t) \} $$

is Euclidean dense in $M$.

*Proof.* Fix $A = U \mathrm{diag}(x)V^* \in M$. Since $Q$ is dense in $S$, pick $x_k \in Q$ with $x_k \to x$. For fixed $U, V$, the map

$$ \phi_{U,V}: \mathbb{R}^n \to \mathbb{C}^{n \times t}, \quad z \mapsto U \mathrm{diag}(z)V^* $$

is continuous, hence $\phi_{U,V}(x_k) \to \phi_{U,V}(x) = A$.

**Lemma 19.** [Cos02, Theorem 4.8] Let $M \subset \mathbb{R}^n$ and $N \subset \mathbb{R}^p$ be semialgebraic smooth submanifolds, and let $f: M \to N$ be a semialgebraic $C^\infty$ mapping. Then the set of critical values of $f$ is a semialgebraic subset of $N$, of dimension $< \mathrm{dim} N$.

**Lemma 20.** Consider a $\Pi_n^\pm$-invariant real algebraic set $S \subseteq \mathbb{R}^n$ and the induced variety $M := \sigma^{-1}(S) \subseteq \mathbb{C}^{n \times t}$ (viewed inside $V^\mathbb{R} \cong \mathbb{R}^{2nt}$). Then there exists a semialgebraic open dense subset $M^\circ \subseteq M^{\mathrm{reg}}$ such that: for every $X \in M^\circ$ and every decomposition $X = U \mathrm{diag}(x)V^*$ with $x \in S^{\mathrm{reg}}$, one has

$$ T_X(M) = \{ UZ_1 \mathrm{diag}(x)V^* + U \mathrm{diag}(x)Z_2^*V^* + U \mathrm{diag}(a)V^*: a \in T_x(S), Z_1, Z_2 \text{ skew-Hermitian} \}. $$

*Proof.* $V^\mathbb{R} \cong \mathbb{R}^{2nt}$, so that $U(n)$ and $U(t)$ are smooth real algebraic submanifolds (defined by polynomial equations in real and imaginary parts). Consider the semialgebraic $C^\infty$ map

$$ \Phi: U(n) \times S^{\mathrm{reg}} \times U(t) \to V^\mathbb{R}, \quad (U, x, V) \mapsto U \mathrm{diag}(x)V^*. $$

Its image is contained in $M$, and it is Euclidean dense in $M$ (Lemma 18).

Apply Lemma 19 to $\Phi$ with target manifold $N := M^{\mathrm{reg}}$: the set of critical values $\mathrm{CritVal}(\Phi) \subseteq M^{\mathrm{reg}}$ is semialgebraic of dimension $< \mathrm{dim} M^{\mathrm{reg}}$. Hence

$$ M^\circ := M^{\mathrm{reg}} \setminus \mathrm{CritVal}(\Phi) $$

is semialgebraic open dense in $M^{\mathrm{reg}}$. For any $X \in M^\circ$ and any $(U, x, V)$ with $\Phi(U, x, V) = X$, the differential $d\Phi_{(U,x,V)}$ is surjective onto $T_X(M)$.

Now compute $d\Phi$. We have

$$ T_U U(n) = \{ UZ_1: Z_1^* = -Z_1 \}, \quad T_V U(t) = \{ VZ_2: Z_2^* = -Z_2 \}, \quad T_x S^{\mathrm{reg}} = T_x S. $$

Therefore for $(UZ_1, a, VZ_2) \in T_U U(n) \times T_x S \times T_V U(t)$,

$$ \begin{aligned} d\Phi_{(U,x,V)}(UZ_1, a, VZ_2) &= UZ_1 \mathrm{diag}(x)V^* + U \mathrm{diag}(a)V^* + U \mathrm{diag}(x)(VZ_2)^* \\ &= UZ_1 \mathrm{diag}(x)V^* + U \mathrm{diag}(a)V^* + U \mathrm{diag}(x)Z_2^*V^*. \end{aligned} $$

Since $d\Phi_{(U,x,V)}$ is surjective, its image equals $T_X(M)$. □
**Lemma 21** (Generic data avoid the bad tangent locus). Let $M \subset V^\mathbb{R}$ be a real algebraic set and let $M^\circ \subset M^{\text{reg}}$ be the open dense subset from Lemma 20. Then there exists a semialgebraic open dense set $\Omega \subset V^\mathbb{R}$ such that for every $Y \in \Omega$, all HD critical points of $Y$ on $M$ lie in $M^\circ$.

*Proof.* For $X \in M^{\text{reg}}$, the HD critical condition is $Y - X \in (T_X M)^{\perp q}$. For fixed $X$, this is an affine space of dimension $\dim V^\mathbb{R} - \dim M$.

Let $B := M^{\text{reg}} \setminus M^\circ$. By Lemma 20, $\dim B < \dim M$. Consider

$$ \mathcal{I} = \{(X, Y) \in B \times V^\mathbb{R} : Y - X \in (T_X M)^{\perp q}\}. $$

Each fiber over $X$ has dimension $\dim V^\mathbb{R} - \dim M$, so

$$ \dim \mathcal{I} \le \dim B + (\dim V^\mathbb{R} - \dim M) < \dim V^\mathbb{R}. $$

Hence its projection to $V^\mathbb{R}$ has dimension $< \dim V^\mathbb{R}$. Outside this set, no critical point lies in $B$. $\square$

**Lemma 22.** Let $<A, B> = \text{Tr}(AB^*)$ be the inner product on $M_n(\mathbb{C})$. Then:

(i) If $\text{Re} <A, Z> = 0$ for all skew Hermitian $Z$, then $A$ is Hermitian.

(ii) If $<A, Z> = 0$ for all skew Hermitian $Z$, then $A = 0$.

*Proof.* Every matrix $A$ decomposes uniquely as

$$ A = H + K, \quad H := \frac{1}{2}(A + A^*) \text{ (Hermitian)}, \quad K := \frac{1}{2}(A - A^*) \text{ (skew Hermitian)}. $$

If $H$ is Hermitian and $K$ is skew Hermitian, then

$$ <H, K> = \text{Tr}(HK^*) = -\text{Tr}(HK). $$

On the other hand,

$$ \overline{<H, K>} = \overline{\text{Tr}(HK^*)} = \text{Tr}((HK^*)^*) = \text{Tr}(KH) = \text{Tr}((HK)^*) = \overline{\text{Tr}(HK)}. $$

Thus $\text{Tr}(HK)$ is purely imaginary, and therefore

$$ (*) \qquad \text{Re} <H, K> = 0. $$

(1) Real-part condition. Assume $\text{Re} <A, Z> = 0$ for all skew Hermitian $Z$. Taking $Z = K$, we obtain

$$ 0 = \text{Re} <A, K> = \text{Re} <H + K, K> = \text{Re} <H, K> + \text{Re} <K, K>. $$

By $(*)$, $\text{Re} <H, K> = 0$, so

$$ 0 = \text{Re} <K, K> = \text{Tr}(KK^*) \ge 0. $$

Hence $K = 0$ and $A = H$ is Hermitian.

(2) Full inner-product condition. Assume $<A, Z> = 0$ for all skew Hermitian $Z$. If $H$ is Hermitian, then $Z = iH$ is skew Hermitian, so

$$ 0 = \langle A, iH \rangle = i \langle A, H \rangle \Rightarrow \langle A, H \rangle = 0 $$

for all Hermitian $H$. Any matrix $B$ can be written as $B = H_1 + iH_2$ with $H_1, H_2$ Hermitian, hence

$$ \langle A, B \rangle = \langle A, H_1 \rangle + i \langle A, H_2 \rangle = 0. $$

Thus $A$ is orthogonal to all matrices, so

$$ 0 = \langle A, A \rangle = \text{tr}(AA^*), $$

which implies $A = 0$. $\square$

**Theorem 23** (Real ED degree of $S$ and HD degree of $M$). Consider a $\Pi_n^\pm$-invariant variety $S \subseteq \mathbb{R}^n$ and the induced real variety $M := \sigma^{-1}(S) \subseteq \mathbb{C}^{n \times t}$. Then for all $Y \in \mathbb{C}^{n \times t}$ in a semialgebraic open dense subset, $Y$ admits an SVD $Y = U \text{diag}(y)V^*$ with $U \in U(n)$, $V \in U(t)$, and $y \in \mathbb{R}^n$. Moreover, the set of Hermitian distance critical points of $Y$ with respect to $M$ is

$$ \{ \text{Udiag}(x)V^* : x \text{ is an ED critical point of } y \text{ with respect to } S \}. $$

In particular, the Hermitian distance degree of $M$ equals the real Euclidean distance degree of $S$:

$$ \text{HDdeg}(M) = \text{REDdeg}(S). $$
Proof. Let $Y \in \mathbb{C}^{n \times t}$ with distinct singular values and let $Y = U \text{diag}(y)V^*$ be an SVD of Y. Let $X$ be an HD critical point of $Y$ with respect to $M$. By Lemma 17 we may assume that $X$ can be written as

$$X = U \text{diag}(x)V^*$$

for some $x \in S$.

We can further assume that $X$ is generic and by Lemma 21 the tangent space $T_X(M)$ at $X = U \text{diag}(x)V^*$ is given in Lemma 20.

We will now show that $x$ is an ED critical point of $y$ with respect to $S$. To see this, observe the inclusion

$$\{U \text{diag}(a)V^*: a \in T_x(S)\} \subseteq T_X(M),$$

and since $y, x$ and $a$ are in $\mathbb{R}^n$ the trace is real and hence we have,

$$0 = \text{Tr}(U \text{diag}(y-x)V^* (U \text{diag}(a)V^*)^*) \quad \text{for any } a \in T_x(S).$$

$$\text{therefore } 0 = \text{Tr}(U \text{ diag}(y-x)V^* \cdot V \text{ diag}(\bar{a}) U^*)$$

$$(y-x)^{\top}\bar{a} = 0 \quad \text{for any } a \in T_x(S),$$

and hence $x$ is an ED critical point of $y$ with respect to $S$. Conversely, suppose $x \in S^{\text{reg}}$ is an ED critical point of $y$ with respect to $S$.

Define now the matrix $X := U \text{diag}(x)V^*$. If moreover $X \in M^\circ$ (the open dense set from Lemma 20), then the tangent space at $X$ has the stated form and is generated by:

(i) $UZ \text{diag}(x)V^*$, with $Z$ skew-Hermitian,

(ii) $U \text{diag}(a)V^*$, where $a \in T_x(S)$,

(iii) $U \text{diag}(x)Z^*V^*$, with $Z$ skew-Hermitian.

We will show for each generator $G$ as in (i) (ii) (iii), we have the $\text{Re}(\text{Tr}((X - Y)G^*))$ vanishes

For (i) and (iii), observe since $x,y$ and $a$ are all in $\mathbb{R}^n$ and $\text{Re}(\text{Tr}(AB^*)) = 0$ for A Hermitian and B skew Hermitian:

$$\begin{align*}
\text{ReTr}((X - Y)(UZ_1\text{diag}(x)V^*)^*) &= \text{ReTr}(\text{diag}(x - y)\text{diag}(x)^*Z_1^*) = 0 \\
(1) \qquad \qquad \qquad \qquad &= \text{ReTr}(\text{diag}(x - y)\text{diag}(x)Z_2) \\
&= \text{ReTr}((X - Y)(U\text{diag}(x)Z_2^*V^*)^*)
\end{align*}$$

In case (ii) we have from the fact $x$ is an ED critical point for $y$ on $S$

$$\text{Tr}((X - Y)(U \text{diag}(a)V^*)^*) = \text{Tr}(U \text{diag}(x - y)\text{diag}(a)^*U^*) = \text{Tr}(\text{diag}(x - y)\text{diag}(a)^*) = 0$$

$$\text{therefore we have the result } \text{HDdeg}(M) = \text{REDdeg}(S). \quad \square$$

This theorem gives us the Hermitian analogue for the main result in [DLOT17].

**Example 24.** [Singular-Value Relation $\{\sigma_1, \sigma_2\} = \{a, a^2\}$] For $2 \times 2$ matrices whose singular values satisfy

$$\{\sigma_1, \sigma_2\} = \{a, a^2\} \quad (a \ge 0),$$

the corresponding absolutely symmetric set is

$$S = \Pi^{\pm} \cdot \{(a, a^2) : a \ge 0\},$$

i.e. the set of all signed permutations of the pair $(a, a^2)$. Equivalently, $S$ may be described as the real algebraic set

$$S = \{(x_1, x_2) \in \mathbb{R}^2 : (x_2 - x_1^2)(x_1 - x_2^2) = 0\},$$

which is invariant under permutations and sign changes of the coordinates.

Let $E_1$ (resp. $E_2$) be the evolute of the parabola $x_2 = x_1^2$ (resp. $x_1 = x_2^2$) (see figure 1). For a generic point $y$, the number of real ED critical points contributed by a given parabola is 1 if $y$ lies on the same side of its evolute as the parabola, and it is 3 if $y$ lies on the other side. Equivalently, if $y$ is outside $E_1$, then $x_2 = x_1^2$ contributes 1 real critical point if $y$ is inside $E_1$, it contributes 3. If $y$ is outside $E_2$, then $x_1 = x_2^2$ contributes 1 real critical point if $y$ is inside $E_2$, it contributes 3.

Hence the total number of real ED critical points of $y$ on $S$ equals 2, 4, or 6, depending on whether $y$ lies inside none, exactly one, or both of the evolutes. By Theorem 23, for a generic matrix with SVD $Y = U \text{diag}(y)V^*$ the Hermitian distance critical points of $Y$ on $M = \sigma^{-1}(S)$ are exactly the matrices
$U \mathrm{diag}(x)V^*$ where $x$ ranges over these real ED critical points of $y$ on $S$. Hence the HD critical count for $Y$ equals the ED critical count for $y$, and it jumps precisely when $y$ crosses the evolutes.

<figure><img src="image_2.png" /><figcaption>FIGURE 1. The absolutely symmetric set $S = \{(x_1, x_2) : (x_2 - x_1^2)(x_1 - x_2^2) = 0\}$ (union of two parabolas) together with the evolutes.</figcaption></figure>

*Example 25 (Determinant-magnitude constraint in $2 \times 2$). Let*

$$M_{\det} := \{A \in \mathbb{C}^{2 \times 2} : |\det A| = 1\},$$

viewed as a real algebraic set in $V^\mathbb{R} \cong \mathbb{R}^8$, equipped with $q = \mathrm{ReTr}(AB^*)$. This set is invariant under the left-right action of $U(2) \times U(2)$ since $|\det(UAV^*)| = |\det A|$.

Its diagonal slice is

$$S = \sigma(M_{\det}) = \{x \in \mathbb{R}^2 : |\det(\mathrm{diag}(x))| = 1\} = \{x \in \mathbb{R}^2 : |x_1 x_2| = 1\} = H_2^+ \cup H_2^-,$$

where $H_2^\pm = \{x_1 x_2 = \pm 1\}$.

Fix $y = (y_1, y_2) \in \mathbb{R}^2$ and parametrize the two branches by

$$H_2^+ : (t, 1/t), \quad H_2^- : (t, -1/t) \quad (t \neq 0).$$

A point on $H_2^\pm$ is ED-critical for $y$ iff $t$ is a real root of

$$q_y^{\pm}(t) := t^4 - y_1 t^3 \pm y_2 t - 1.$$

Let $D^\pm(y)$ be the discriminant of $q_y^\pm(t)$ as a polynomial in $t$:

$$D^+(y) = -256 + 192y_1y_2 + 6y_1^2y_2^2 + 4y_1^3y_2^3 - 27y_1^4 - 27y_2^4,$$

$$D^-(y) = -256 - 192y_1y_2 + 6y_1^2y_2^2 - 4y_1^3y_2^3 - 27y_1^4 - 27y_2^4.$$

For $y$ not on the semialgebraic curve $\{D^+(y)D^-(y) = 0\}$, the number of real ED-critical points of $y$ on $S = H_2^+ \cup H_2^-$ is

$$\mathrm{REDdeg}(S : y) = \begin{cases} 6, & \text{if } D^+(y) > 0 \text{ or } D^-(y) > 0, \\ 4, & \text{if } D^+(y) < 0 \text{ and } D^-(y) < 0, \end{cases}$$

see [DLT15, Example 4.4].
Now let $Y \in \mathbb{C}^{2 \times 2}$ be generic and write an SVD $Y = U \text{diag}(y)V^*$. By Theorem 23, the HD-critical points of $Y$ on $M_{\text{det}}$ are exactly

$$ \{ U \text{diag}(x)V^* : x \in S \text{ is ED-critical for } y \}, $$

so the chamber-wise Hermitian critical count for $M_{\text{det}}$ is $\{4, 6\}$, and it jumps precisely when $y$ crosses $\{D^{+}(y)D^{-}(y) = 0\}$.

*Example 26 (Unit Schatten d-sphere).* Fix an even integer $d \ge 2$. For $X \in \mathbb{C}^{n \times t}$, define the Schatten d-norm

$$ \|X\|_d := \left( \sum_{i=1}^{n} \sigma_i (X)^d \right)^{1/d}, $$

where $\sigma_1(X), \dots, \sigma_n(X)$ are the singular values of $X$. Consider the (Unitary-invariant) unit sphere

$$ \mathcal{F}_{n,t,d} := \{ X \in \mathbb{C}^{n \times t} : \|X\|_d = 1 \}. $$

Since $d$ is even, $\mathcal{F}_{n,t,d}$ is a real algebraic set in $V \mathbb{R} \cong \mathbb{R}^{2nt}$.

Its singular-value image is the affine Fermat hypersurface

$$ F_{n,d} := \{ x \in \mathbb{R}^n : \sum_{i=1}^{n} x_i^d = 1 \}, $$

which is absolutely symmetric. Hence $\mathcal{F}_{n,t,d} = \sigma^{-1}(F_{n,d})$.

Let $Y \in \mathbb{C}^{n \times t}$ be generic and write an SVD $Y = U \text{diag}(y)V^*$ with $y \in \mathbb{R}^n$. By Theorem 23, the Hermitian distance critical points of $Y$ on $\mathcal{F}_{n,t,d}$ are precisely the matrices $U \text{diag}(x)V^*$ where $x$ ranges over the real ED critical points of $y$ on $F_{n,d}$. In particular,

$$ \text{HDdeg}(\mathcal{F}_{n,t,d}) = \text{REDdeg}(F_{n,d}). $$

In the case $n = 2$, write $F_{2,d} = \{(x_1, x_2) \in \mathbb{R}^2 : x_1^d + x_2^d = 1\}$. A point $x \in F_{2,d}$ is a real ED critical point of $y = (y_1, y_2)$ if and only if it lies on the curve $\gamma_{d,y}$.

$$ \gamma_{d,y} := \{ x \in \mathbb{R}^2 : x_1^{d-1}(x_2 - y_2) = x_2^{d-1}(x_1 - y_1) \}. $$

see [DLT15, Example 4.3].

For $d = 2$ one has $\#(F_{2,2} \cap \gamma_{2,y}) = 2$ for generic $y$ (the circle case). For every even $d \ge 4$, there exists a nonempty open set of $y \in \mathbb{R}^2$ for which $\#(F_{2,d} \cap \gamma_{d,y}) = 8$, hence $\text{HDdeg}(\mathcal{F}_{2,t,d}) = 8$ in that generic chamber.

*Remark 27.* Consider the unitary group $U(n)$ then the absolute symmetric set $S$ is given by $(\pm 1, \dots, \pm 1)$, since each point is critical in the variety the ED degree of $S$ is $2^n$. Therefore the HD degree of $U(n)$ is the same. If $A$ a general matrix has an SVD given by $U \Sigma V$ then the corresponding critical points are obtained at $U \text{diag}(x)V$ such that $x \in S$.

Let $A = \begin{bmatrix} i & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$. One of the SVD of $A$ is given by $A = U \Sigma V^*$, where $\Sigma = \text{diag}(1, 2, 3)$,

$U = \text{diag}(i, 1, 1)$, and $V = I$. Then the critical points should admit a simultaneous decomposition and therefore should be $U \text{diag}(\pm 1, \pm 1, \pm 1)$. Note one of the critical points is $U$ itself and from 20, $U \text{diag}(i, -2i, 3i)$ is an element of $T_U M$. Now if you compute $\text{Tr}((A - U)(U \text{diag}(i, -2i, 3i))^*) = -4i$, so the imaginary part remains. This is due to the fact that the tangent space is real linear and the unitary group is a not a complex submanifold.

*Remark 28.* Notice if $M$ is a complex sub manifold of $\mathbb{C}^{n \times t}$ which is invariant under unitary actions and $X = U \text{diag}(x)V$ is a critical point for $Y = U \text{diag}(y)V$ a data point, then in particular from (1), Lemma 22 and Lemma 8 we have $\text{diag}(x - y)\text{diag}(x) = 0$.

Let $V = \mathbb{C}^{m \times n}$ with $m \le n$ and Hermitian inner product $\langle A, B \rangle = \text{Tr}(AB^*)$. For $1 \le r \le m$ let

$$ X_r := \{ A \in \mathbb{C}^{m \times n} : \text{rank}(A) \le r \}. $$

Fix $U \in \mathbb{C}^{m \times n}$ and write a singular value decomposition

$$ U = T_1 \Sigma T_2, \quad \Sigma = \text{diag}(\sigma_1, \dots, \sigma_m), \quad \sigma_1 \ge \dots \ge \sigma_m \ge 0, $$

with $T_1 \in U(m)$ and $T_2 \in U(n)$. For $1 \le j \le m$ let $\Sigma_j$ denote the diagonal matrix with $\sigma_j$ in the $j$-th diagonal position and zeros elsewhere.
Assume $U$ is generic (e.g. $\sigma_1 > \cdots > \sigma_m$). Then the Hermitian critical points of the squared distance function $d_U(A) = \|U - A\|^2$ on $X_r$ are precisely the matrices

$$T_1(\Sigma_{i_1} + \cdots + \Sigma_{i_r})T_2, \quad 1 \le i_1 < \cdots < i_r \le m.$$

Consequently, the Hermitian distance polynomial $HDpoly_{X,U}$ as in [Fur25], i.e. the Euclidean distance polynomial of the real variety with respect to the real inner product $\mathrm{Re}\langle\cdot, \cdot\rangle$ induced by the standard Hermitian form.

$$HDpoly_{X_r,U}(t^2) = \prod_{1 \le i_1 < \cdots < i_r \le m} \left( t^2 - \sum_{j \notin \{i_1, \dots, i_r\}} \sigma_j(U)^2 \right).$$

In particular, for the corank one determinantal variety $X_{m-1}$ we obtain

$$HDpoly_{X_{m-1},U}(t^2) = \prod_{k=1}^{m} (t^2 - \sigma_k(U)^2) = \det(t^2 I_m - UU^*).$$

This is the Hermitian analogue of [OS20, Theorem 8.1]
## 4. SLICING THEOREM FOR HERMITIAN DISTANCE DEGREE

Section 3 gives an explicit description of Hermitian critical points and their critical values via the singular-value reduction. In this section we explain how the same reduction also follows abstractly from a slicing principle of Bik-Draisma. We state the result for the Hermitian distance degree, paralleling the Euclidean distance degree result in [AJ18].

**Theorem 29.** [AJ18][Bik-Draisma] Let $V$ be a finite dimensional vector space over $\mathbb{C}$ equipped with a Euclidean product $\langle\cdot, \cdot\rangle$. Let $G$ be a complex algebraic group with an orthogonal representation $O(V)$. Suppose that $V$ has a linear subspace $V_0$ such that, for sufficiently general $v_0 \in V_0$, the space $V$ is the orthogonal direct sum of $V_0$ and the tangent space $T_{v_0}(G \cdot v_0)$ to its $G$-orbit. Let $X$ be a $G$-stable closed subvariety of $V$. Set $X_0 := X \cap V_0$ and suppose that $G \cdot X_0$ is dense in $X$. Then the ED degree of $X$ in $V$ equals the ED degree of $X_0$ in $V_0$.

Let $V$ be a finite-dimensional complex vector space with Hermitian inner product $\langle\cdot, \cdot\rangle$, and let

$$q := \mathrm{Re}\langle\cdot, \cdot\rangle$$

be the associated real Euclidean inner product on the realification $V^R$. For a real algebraic subset $X \subseteq V^R$ (in particular, for a complex variety viewed as a real set), the critical points of the squared Hermitian distance function are exactly the real Euclidean distance critical points in the inner-product space $(V^R, q)$. Hence the relevant enumerative invariant is a chamberwise real critical count.

If $G \to U(V)$ is a unitary representation, then $G$ acts on $V^R$ by $q$-orthogonal transformations. Therefore the slicing method of Bik-Draisma [AJ18] applies in the real inner-product space $(V^R, q)$; we include the argument for completeness, emphasizing that tangent spaces are generally only real linear in the Hermitian distance problem. Throughout, $\mathrm{HDdeg}(X)$ denotes the number of real Euclidean distance critical points of a sufficiently general data point in $V^R$ with respect to $q$.

**Theorem 30** (Hermitian slicing Theorem). Let $V$ be a finite-dimensional complex vector space with Hermitian inner product $\langle\cdot, \cdot\rangle$, and let $q = \mathrm{Re}\langle\cdot, \cdot\rangle$ be the associated real Euclidean inner product on $V^R$. Let $G \to U(V)$ be a unitary representation (hence an orthogonal representation on $(V^R, q)$).

Suppose that $V^R$ has a real linear subspace $V_0^R$ such that, for sufficiently general $v_0 \in V_0^R$, the space $V^R$ is the $q$-orthogonal direct sum of $V_0^R$ and the tangent space $T_{v_0}(G \cdot v_0)$. Let $X$ be a $G$-stable real subvariety of $V^R$. Set $X_0 := X \cap V_0^R$ and suppose that $G \cdot X_0$ is dense in $X$. Then $\mathrm{HDdeg}_{V^R}(X) = \mathbb{R}\mathrm{EDdeg}_{V_0^R}(X_0)$.

**Lemma 31.** [AJ18](Lemma 16) Critical points are equivariant: if $x$ is a critical point for $u$, then $g \cdot x$ is a critical point for $g \cdot u$.

**Lemma 32.** [AJ18](Lemma 15) The set $GV_0$ is dense in $V$

**Lemma 33.** [AJ18](Lemma 17) A sufficiently general $x_0 \in X_0$ lies in $X^{\mathrm{reg}}$ and $X_0^{\mathrm{reg}}$

**Lemma 34.** A sufficiently general $x_0 \in X_0$ satisfies

$$T_{x_0}X = T_{x_0}X_0 + T_{x_0}(G \cdot x_0).$$

*Proof.* Consider the action map (on real loci)

$$m: G \times X_0 \to X, \quad (g, x) \mapsto g \cdot x.$$

By assumption $G \cdot X_0$ is dense in $X$, so the image of $m$ is Euclidean dense in $X$. Restrict $m$ to the smooth loci:

$$m^{\mathrm{reg}}: G \times X_0^{\mathrm{reg}} \to X^{\mathrm{reg}},$$

which is a semialgebraic $C^\infty$ map between semialgebraic smooth manifolds.

Apply Lemma 19 to $m^{\mathrm{reg}}$. The set of critical values in $X^{\mathrm{reg}}$ is semialgebraic of dimension $< \dim X^{\mathrm{reg}} = \dim X$. Hence the set of regular values is dense in $X^{\mathrm{reg}}$. Pick a point $x \in X^{\mathrm{reg}}$ that is a regular value of $m^{\mathrm{reg}}$ and lies in the dense set $G \cdot X_0^{\mathrm{reg}}$. Choose $(g, x_0) \in G \times X_0^{\mathrm{reg}}$ with $g \cdot x_0 = x$. Since $x$ is a regular value, the differential $dm_{(g,x_0)}$ is surjective onto $T_x X$.

For $\xi \in T_g G$ and $v \in T_{x_0} X_0$,

$$dm_{(g,x_0)}(\xi, v) = \xi \cdot x + g \cdot v,$$

so

$$\mathrm{Im}(dm_{(g,x_0)}) = T_x(G \cdot x) + g \cdot T_{x_0} X_0.$$
Surjectivity gives

$$T_x X = T_x(G \cdot x) + g \cdot T_{x_0} X_0.$$

Apply $g^{-1}$ and use $g^{-1} \cdot T_x X = T_{x_0} X$ and $g^{-1} \cdot T_x(G \cdot x) = T_{x_0}(G \cdot x_0)$ to obtain

$$T_{x_0} X = T_{x_0}(G \cdot x_0) + T_{x_0} X_0.$$

Finally, the set of such $x_0$ is dense in $X_0$, so the equality holds for sufficiently general $x_0 \in X_0$. $\square$

**Lemma 35.** Let $v_0 \in V_0^\mathbb{R}$ then any $\mathbb{R}$ED critical point on $X_0$ for $v_0$ is an HD critical point on $X$ for $v_0$

*Proof.* Let $x_0$ be a real ED critical point for $v_0$ in $X_0$ then $v_0 - x_0 \perp T_{x_0} X_0$ and since $v_0 - x_0 \in V_0$ which is orthogonal to $T_{x_0} G x_0$ therefore by Lemma 34 we have $v_0 - x_0 \perp T_{x_0} X$ $\square$

**Lemma 36.** Let $v_0 \in V_0$ be sufficiently general. Then any HD critical point on $X$ for $v_0$ is an $\mathbb{R}$ED critical point on $X_0$ for $v_0$.

*Proof.* Let $x$ be an HD critical point for $v_0$ on $X$. Then $\mathrm{Re} \langle x - v_0, T_x X \rangle = 0$ and also we have that there exists $x_0$ such that $T_x X = g \cdot T_{x_0} X = g \cdot T_{x_0} X_0 + g \cdot T_{x_0}(G x_0) = T_x X_0^g + T_x(G x_0)$. Since $T_x(G \cdot x_0) = g \cdot (g \cdot x_0) = \{g \cdot (\xi \cdot x_0) \mid \xi \in g\} = g \cdot x$ we have $x - v_0 \perp g \cdot x$. Then we have $v_0 \perp g \cdot x$ and since $G$ is a unitary representation we have $\mathrm{Re} \langle g \cdot x, v_0 \rangle = -\mathrm{Re} \langle x, g \cdot v_0 \rangle = 0$. Since $v_0$ is sufficiently general in $V_0^\mathbb{R}$, the vector space $V^\mathbb{R}$ is the orthogonal direct sum of $V_0^\mathbb{R}$ and $T_{v_0} G v_0$, and therefore $x \in V_0^\mathbb{R}$. Since $v_0 - x \perp T_x X \supseteq T_x X_0$, we find that $x \in X_0$ is a real ED critical point for $v_0$. $\square$

By 31 and 32 we may assume that the sufficiently general point on $V^\mathbb{R}$ is in fact a sufficiently general point $v_0$ on $V_0^\mathbb{R}$. The previous two lemmas now tell us that the HD critical points for $v_0$ on $X$ and on $X_0$ are the same. **Hence the HD degree of $X$ in $V^\mathbb{R}$ is the $\mathbb{R}$ED degree of $X_0$ in $V_0^\mathbb{R}$**.

We now explain how Theorem 23 fits into the general slicing principle of Theorem 30. Throughout we assume $n \le t$ and we consider the left right unitary action

$$G := U(n) \times U(t), V := \mathbb{C}^{n \times t}, \quad (U, V) \cdot A := UAV^*.$$

Define the (real) diagonal slice

$$V_0^\mathbb{R} := \{ \mathrm{diag}(x) \in \mathbb{C}^{n \times t} : x \in \mathbb{R}^n \},$$

Let $v_0 = \mathrm{diag}(y) \in V_0^\mathbb{R}$ with $y_1, \dots, y_n$ nonzero and pairwise distinct. Then the tangent space to the $G$ orbit at $v_0$ is

$$T_{v_0}(G \cdot v_0) = \{ Z_1 \mathrm{diag}(y) + \mathrm{diag}(y) Z_2^* : Z_1 \in \mathfrak{u}(n), Z_2 \in \mathfrak{u}(t) \}.$$

Moreover $V^\mathbb{R}$ splits $q$ orthogonally as

$$(2) \qquad V^\mathbb{R} = V_0^\mathbb{R} \oplus_{\mathbb{R}} T_{v_0}(G \cdot v_0).$$

Indeed, for any $a \in \mathbb{R}^n$ and any $Z_1 \in \mathfrak{u}(n)$,

$$q(\mathrm{diag}(a), Z_1 \mathrm{diag}(y)) = \mathrm{Re} \mathrm{Tr}(\mathrm{diag}(a)(Z_1 \mathrm{diag}(y))^*) = \mathrm{Re} \mathrm{Tr}(\mathrm{diag}(a)\mathrm{diag}(y)Z_1^*) = 0,$$

since $\mathrm{diag}(a)\mathrm{diag}(y)$ is Hermitian and $Z_1^*$ is skew Hermitian, hence the trace is purely imaginary. The same argument shows $q(\mathrm{diag}(a), \mathrm{diag}(y)Z_2^*) = 0$ for all $Z_2 \in \mathfrak{u}(t)$. Therefore

$$V_0^\mathbb{R} \subseteq (T_{v_0}(G \cdot v_0))^{\perp q}.$$

To prove (2), it remains to show the reverse inclusion. Let $A \in V^\mathbb{R}$ satisfy $q(A, T_{v_0}(G \cdot v_0)) = 0$. Since

$$T_{v_0}(G \cdot v_0) = \{ Z_1 \mathrm{diag}(y) + \mathrm{diag}(y) Z_2^* : Z_1 \in \mathfrak{u}(n), Z_2 \in \mathfrak{u}(t) \},$$

we obtain for all $Z_1 \in \mathfrak{u}(n)$ and $Z_2 \in \mathfrak{u}(t)$ the two conditions

$$0 = q(A, Z_1 \mathrm{diag}(y)) = \mathrm{Re} \mathrm{Tr}(A \mathrm{diag}(y)^* Z_1^*), \quad 0 = q(A, \mathrm{diag}(y) Z_2^*) = \mathrm{Re} \mathrm{Tr}(\mathrm{diag}(y)^* A Z_2).$$

By Lemma 22(1), the first family implies that $A \mathrm{diag}(y)^*$ is Hermitian, and the second implies that $\mathrm{diag}(y)^* A$ is Hermitian. Since $v_0 = \mathrm{diag}(y)$ has nonzero diagonal entries with pairwise distinct $|y_i|^2$, Lemma 16 yields that $A$ must be a real diagonal matrix (in particular $A \in V_0^\mathbb{R}$). Hence

$$(T_{v_0}(G \cdot v_0))^{\perp q} \subseteq V_0^\mathbb{R}.$$
Therefore $(T_{v_0}(G \cdot v_0))^{\perp q} = V_0^{\mathbb{R}}$, and the $q$ orthogonal direct sum decomposition (2) follows.

Now let $M \subseteq \mathbb{C}^{n \times t}$ be a $G$ stable real algebraic variety (i.e. a unitary invariant matrix variety), and define

$$S := \sigma(M) = \{x \in \mathbb{R}^n : \operatorname{diag}(x) \in M\}.$$

Set $X := M$ and $X_0 := X \cap V_0^{\mathbb{R}}$. By construction we have

$$X_0 = \{\operatorname{diag}(x) : x \in S\},$$

so under the identification $V_0^{\mathbb{R}} \cong \mathbb{R}^n$ given by $x \leftrightarrow \operatorname{diag}(x)$, the slice $X_0$ corresponds exactly to $S$.

Finally, since $M$ is unitary invariant, every $A \in M$ admits an SVD $A = U \operatorname{diag}(\sigma(A)) V^*$ with $U \in U(n)$, $V \in U(t)$, and $\operatorname{diag}(\sigma(A)) \in M$. Hence

$$M = G \cdot X_0,$$

so $G \cdot X_0$ is dense in $X = M$. Therefore all hypotheses of Theorem 30 are satisfied (for generic $v_0 \in V_0^{\mathbb{R}}$), and we obtain

$$\mathrm{HDdeg}_V(M) = \mathrm{REDdeg}_{V_0^{\mathbb{R}}}(X_0).$$

*Remark 37.* This is the same equality as in Theorem 23, since $\mathrm{REDdeg}_{V_0^{\mathbb{R}}}(X_0)$ coincides with the real Euclidean distance degree of $S$. Indeed, on $V_0^{\mathbb{R}}$ (the space of real diagonal matrices), the form $q$ restricts to the standard Euclidean inner product: for $A, B \in V_0^{\mathbb{R}}$ we have

$$\mathrm{Re}(\mathrm{Tr}(AB^*)) = \mathrm{Tr}(AB^t),$$

so the associated squared norm is $A \mapsto \mathrm{Tr}(AA^t)$.

We also show how Theorem 30 can be used to study Critical points of rank one Hermitian matrices in the next example

Let $G = U(n)$ act on $V$ by conjugation,

$$g \cdot A = gAg^*.$$

This action is unitary since $\mathrm{Tr}(gAg^*(gBg^*)^*) = \mathrm{Tr}(AB^*)$, so the real part of $q$ defines a $G$ invariant Euclidean structure on $V^{\mathbb{R}}$.

Define

$$V_0 = \{\operatorname{diag}(\lambda_1, \dots, \lambda_n) : \lambda_i \in \mathbb{C}\},$$

the space of diagonal complex matrices. For a sufficiently general element $v_0 = \operatorname{diag}(\lambda_1, \dots, \lambda_n) \in V_0$ with $\lambda_i \neq \lambda_j$ for $i \neq j$, the tangent space to the $G$ orbit is

$$T_{v_0}(Gv_0) = \{[H, v_0] : H \in \mathfrak{u}(n)\} = \{A \in \mathbb{C}^{n \times n} : A \text{ is off diagonal}\}.$$

Since diagonal and off diagonal matrices are orthogonal with respect to $\mathrm{Re}(q)$, we have an orthogonal direct sum decomposition

$$V^{\mathbb{R}} = V_0^{\mathbb{R}} \oplus_{\mathbb{R}} T_{v_0}(Gv_0).$$

Let $X \subset V$ be the variety of rank one Hermitian matrices,

This is a closed $G$ stable subvariety of $V$. Its intersection with the slice $V_0$ is

$$X_0 = X \cap V_0^{\mathbb{R}} = \{\operatorname{diag}(0, \dots, 0, \lambda, 0, \dots, 0) : \lambda \in \mathbb{R}\},$$

the union of the coordinate axes in $V_0$.

Every rank one Hermitian matrix is unitarily diagonalizable, hence for each $X = vv^* \in X$ there exists $g \in U(n)$ such that $gXg^* \in X_0$. Therefore,

$$G \cdot X_0 = X,$$

so $GX_0$ is dense in $X$.

All hypotheses of the Theorem are thus satisfied, and we conclude that

$$\mathrm{HDdeg}_V(X) = \mathrm{REDdeg}_{V_0^{\mathbb{R}}}(X_0).$$
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Email address: Nikhil.ken@unifi.it