A nine-dimensional counterexample to Borsuk’s covering assertion

Proof. The formula uu^{\mathsf T} is unchanged when u is replaced by -u, so it gives a continuous map on projective space. Its range as an operator is exactly x, which proves injectivity. A continuous injection from the compact space \mathbb P(V) into a Hausdorff space is a homeomorphism onto its image. Every P_x has trace one, and the trace-one hyperplane has codimension one in \operatorname{Sym}(V). Finally, \operatorname{tr}(P_x^2)=\operatorname{tr}(P_y^2)=1 and \operatorname{tr}(P_xP_y)=\langle u,v\rangle^2, proving (2.1). Orthogonal lines exist because k\geq2.□

Proof. Pad the cover by empty sets to write it as A_1,\ldots,A_m. For every nonempty A_i, set d_i=\operatorname{diam}(A_i) and \varepsilon_i=(\sqrt2-d_i)/4>0, and define the relatively open set

Set U_i=\varnothing if A_i=\varnothing. The sets U_i cover X_k. For any P,Q\in U_i, points of A_i within \varepsilon_i of P,Q and the triangle inequality give

In particular \operatorname{diam}(U_i)\leq(d_i+\sqrt2)/2<\sqrt2.

Pull these sets back to a finite open cover V_i of \mathbb P(V). A smooth partition of unity subordinate to this cover [15, Theorem 2.23] gives smooth nonnegative functions f_i, with sum one and f_i(x)>0 only if x\in V_i. Empty sets receive the zero function. If i belonged to both S(x) and S(y) for orthogonal lines x,y, then P_x,P_y\in U_i would be at distance \sqrt2 by Lemma 2.1, a contradiction.□

Proof. The positive square root depends continuously on a positive semidefinite matrix. For example, approximate the square-root function uniformly by polynomials on a compact interval containing the spectra under consideration, and apply those polynomials to the matrices. Thus the integrand in (3.1) is jointly continuous in (Q,u); integration over the compact sphere proves continuity of E_V. Its homogeneity and nonnegativity follow from the corresponding properties of H. Since \sum_i H(v)_i=\|v\|^2 and \int uu^{\mathsf T}\,d\mu_V(u)=k^{-1}I, we have

Every nonzero Q^{1/2}u lies in \operatorname{ran}Q. Conversely, every line in that range is spanned by some Q^{1/2}u with \|u\|=1, because Q^{1/2} is invertible on its range. If f_i is positive on such a line, the ith integrand is positive at that u, and hence on a nonempty open set of the sphere. Such a set has positive measure. This proves (3.2) in both directions. For Q=P_x, every nonzero integrand is a scalar multiple of f(x), and the scalars integrate to \operatorname{tr}P_x=1. Thus E_V(P_x)=f(x).

To prove the subspace identity, write r=\dim W and let \pi_W be orthogonal projection. The finite measure on \mathbb S(W) defined by

is invariant under all orthogonal transformations of W. Its total mass is \int\|\pi_Wu\|^2\,d\mu_V(u)=r/k, so it equals (r/k)\mu_W. Because Q vanishes on W^\perp and H is homogeneous of degree two, this gives

as claimed.

Finally choose, locally in the parameters, a smooth orthonormal frame T:\mathbb R^r\to W. Write Q|_W in that frame as the positive definite matrix C. The subspace formula becomes

The matrix square root is smooth on positive definite matrices, and TC^{1/2}w never vanishes. The function H is smooth away from zero. Differentiating this expression over the compact sphere proves the asserted local smoothness.□

Proof. Continuity and homogeneity follow from Lemma 3.1 and the continuous formulas for A_\pm. Replacing A by -A exchanges A_+ and A_-, so F is odd. Any line in \operatorname{ran}A_+ is orthogonal to every line in \operatorname{ran}A_-. Admissibility and (3.2) therefore imply that E_V(A_+) and E_V(A_-) have disjoint positive coordinates. There is no cancellation in their difference. This proves (3.5) and, by summing coordinates, (3.4).

If \ker A contains a line x, choose i\in S(x), which is possible because \sum_i f_i(x)=1. The line x is orthogonal to both spectral ranges. Thus label i is absent from both unions in (3.5), so F(A)_i=0. This proves the nonsingularity assertion. If all coordinates of F(A) are negative, then E_V(A_+)=0 by disjointness of the two positive supports. Hence \operatorname{tr}A_+=0, so A_+=0, and nonsingularity makes A negative definite. Oddness gives the positive case.

For local smoothness, fix nonsingular A. In a neighborhood of A, the positive and negative spectral subspaces have constant dimensions and depend smoothly on the matrix. Explicitly their projections are

which are smooth there because A^2 is positive definite. Each projection admits a local smooth orthonormal frame: project a fixed basis for its range and apply Gram–Schmidt after shrinking the neighborhood. In these frames, A_+ and A_- restrict to positive definite operators on their respective ranges. The last assertion of Lemma 3.1 proves smoothness of E_V(A_+) and E_V(A_-), with a zero term if the corresponding range has dimension zero. This proves smoothness of F.□

Proof. An odd map S^n\to S^d requires n\le d, and an odd self-map of S^n has odd degree; see [8, Proposition 2B.6 and Corollary 2B.7]. Applying these facts to F proves the bound and the degree assertion. A map of nonzero degree cannot omit a point of its target sphere, since the complement of a point is contractible.

Suppose m=N_k and fix distinct labels i,j. Choose a vector in \Sigma_m with every coordinate nonzero, with precisely the ith and jth coordinates positive. At a preimage A under F, every line in the positive spectral subspace of A uses only these two labels. That subspace cannot have dimension at least two: restriction to a two-dimensional subspace would contradict the bound N_2=3 just proved. Its dimension is therefore one. The exact spectral-support property of F says that this line has both labels i,j in its support, so \{i,j\} is an edge.□

Proof. The inverse function theorem makes every point of the fiber isolated. The fiber is closed in compact M, hence finite. By excision,

The fundamental class has coordinate one in each summand. A local diffeomorphism sends each local generator to the generator at y. Naturality of relative homology now gives the assertion. This argument uses no smoothness away from the selected fiber.□

Proof. The edge assertion is Proposition 4.1. We prove the remaining assertions by examining an arbitrary facet I. Set

Since I is a maximal support face, there is a line x_0 with S(x_0)=I. Every line in x_0^\perp uses only labels in J. Proposition 4.1, applied in x_0^\perp, gives

A compact level set over one affine chart. Define

Maximality of I implies S(x)=I for all x\in U_I. Thus the coordinate map f_I takes U_I smoothly to the open simplex \mathring\Delta_I on I. Moreover, U_I lies in the affine projective chart

because a line orthogonal to x_0 cannot use any label in I. Choose a regular value r\in\mathring\Delta_I of f_I, using Sard’s Theorem [15, Theorem 6.10], and put

For now L is allowed to be empty. If \widetilde r denotes r extended by zero on J, then L=f^{-1}(\widetilde r): positivity of the coordinates of r puts every such preimage in U_I. Hence L is compact. It is a smooth manifold without boundary [15, Corollary 5.14] of dimension

If s<k, the level-set dimension in (4.2) is positive. We will construct a map on a sphere bundle over L whose regular preimages have odd cardinality, by comparison with F. The affine-chart trivialization will force the induced map on the projective bundle to have degree zero when \dim L>0, contradicting that count. When s=k, the same count will determine the coefficient of I in the desired cycle.

For x\in U_I, admissibility ensures that f on \mathbb P(x^\perp) uses only labels in J. Apply the extension construction in this (k-1)-dimensional space to obtain an odd map

The subspace-restriction and continuity properties of the extension make these maps a continuous map on the corresponding sphere bundle. Restrict that bundle to L and write

The bundle x\mapsto x^\perp is trivial on the affine chart above. Indeed, choose the unit representative u of x having positive inner product with a fixed unit representative of x_0. Orthogonal projection from x_0^\perp onto u^\perp is invertible and varies smoothly with x; orthonormalizing its images of a fixed basis gives a smooth orthonormal trivialization. The induced trivialization of symmetric operators, followed by radial change from trace norm to a Euclidean norm, gives

Equip M with the smooth structure given by the Euclidean sphere bundle. The trace-norm parametrization is smooth near every definite B, which is all that will be needed. In particular,

Odd parity on the bundle. Consider the open negative orthant of \Sigma_a, where all J coordinates are strictly negative. At a preimage of this orthant, B has no positive spectral subspace. It has no kernel either: the exact support property says that all J labels occur on its range, leaving no label for a kernel line. Thus B is negative definite on x^\perp. Consequently G is smooth on the preimage of the negative orthant. Sard’s Theorem allows us to choose a regular value g in that orthant. Its preimage is finite by compactness, including the possibility of an empty preimage.

Fix 0<\lambda<1 and form the full-support vector

with the two coordinate blocks indexed by I and J. We claim that F^{-1}(y) corresponds exactly to G^{-1}(g). Every A\in F^{-1}(y) is nonsingular by the full-support property. All lines in its nonzero positive spectral subspace use only labels in I. That subspace must be one-dimensional: a subspace of dimension at least two intersects the hyperplane x_0^\perp, where no label in I is available. Call the resulting positive line x. The positive output block has every I coordinate nonzero, so S(x)=I and x\in U_I. Since the sum of the positive coordinates of F(A) is \lambda, the positive eigenvalue is \lambda. Therefore

The subspace formula for the extension now gives f_I(x)=r and G_x(B)=g. Conversely, every (x,B)\in G^{-1}(g) gives a preimage by (4.5).

The value y is regular in the local sense of Lemma 4.2. To see this precisely, near a preimage use (\lambda',x',B') as coordinates for symmetric operators of signature (1,k-1), with B' negative definite and normalized on x'^\perp. The single positive eigenvalue and its line vary smoothly, so these are smooth coordinates on the trace-norm sphere near that preimage. In the target orthant, use the positive mass followed by the two normalized coordinate blocks. In these coordinates F is

The first derivative block is the identity. The second is surjective because r is regular for f_I. On the kernel of these two blocks, the third block is the derivative of G|_M, which is surjective because g is regular. Thus the full derivative is surjective, and is an isomorphism since source and target both have dimension m-1. By the odd degree of F and Lemma 4.2,

This proves, in particular, that L is nonempty.

The double cover forces maximal supports to have size k. The oddness of G in the sphere variable gives a quotient map

Write w\in H^1(\mathbb{RP}^{a-1}) and u\in H^1(\mathbb{RP}^{h-1}) for the first Stiefel–Whitney classes of the real line bundles associated with their antipodal double covers. Equivariance identifies the pullback of the target sphere cover with

Indeed, the identification sends (x,z) to ((x,[z]),G(x,z)) and is an isomorphism on each two-point fiber. This cover isomorphism induces an isomorphism of the associated real line bundles. Their first Stiefel–Whitney classes are natural under pullback [9, Theorem 3.1(a)]. The source cover is pulled back from the second factor, so

There is no contribution from H^1(L) in this identity.

Suppose a>h. The projective-space cohomology calculation [8, Theorem 3.19] gives

Both source and target of \overline G are closed manifolds of dimension a-1, and w^{a-1} is the target’s top cohomology generator. Evaluation on the total fundamental class therefore gives mod-two degree zero. On the other hand, [g] is a regular value in local quotient coordinates, and every preimage of [g] has exactly one representative mapping to g. Its preimage count is therefore the odd number in (4.7), a contradiction. Together with (4.1), this proves a=h and s=k for every facet.

The coefficients of the facet cycle. Now \dim L=0, so L is a finite set. For each x\in L, the map G_x is odd between spheres of the same dimension h-1. Regularity of g for G implies its regularity for each such fiber map. Each fiber therefore contributes an odd number to (4.7), and hence

Every facet has k vertices, so K has dimension k-1. Using the identification of simplicial and singular homology [8, Theorem 2.27], consider the homology class

The interior point \widetilde r of the facet I has a neighborhood in |K| contained in that facet. Passing to H_{k-1}(|K|,|K|\setminus\{\widetilde r\}) reads the coefficient of I in a simplicial top cycle. By the regularity of r, Lemma 4.2 and (4.9), this coefficient is one. The same holds for every facet. Since a (k-1)-dimensional complex has no simplicial k-chains, its top homology classes are exactly its top cycles. Thus f_*[\mathbb P(V)] is represented by the sum of all facets. Its boundary vanishes. Each codimension-one face consequently occurs in an even number of facets, and purity makes that number positive.□

Proof. By Theorem 4.3, the complex is pure of dimension two, has all fifteen edges, and each edge belongs to a positive even number of triangles. Two complementary triples cannot both be triangles: choose witness lines x,y\subset\mathbb R^3 whose supports contain the respective triples. A line perpendicular to both exists and, by admissibility, could use none of the six labels.

There are ten complementary pairs of triples, so there are at most ten triangles. Counting incidences between triangles and edges gives at least 2\cdot15/3=10 triangles. Equality holds throughout: there is one triangle in each complementary pair, and every edge occurs twice.□

Proof. Every vertex of the link at a has degree two, since the corresponding pair with a lies in exactly two triangles. A simple graph in which every vertex has degree two is a disjoint union of cycles of length at least three. On five vertices it must therefore be a single five-cycle.

Let A\subset C have four elements and choose a\in A. A five-cycle has no independent set of size three: if three vertices were independent, the three gaps following them around the cycle would each contain another vertex, requiring at least six vertices. Thus the other three elements of A contain an edge of the link at a, giving a triangle in A. If all four triples in A were triangles, the link at a would contain the three-cycle on A\setminus\{a\}, which a five-cycle cannot contain.

For the third assertion, write C=\{a,b,c,d,e,f\}, where the two triangles containing \{a,b\} are \{a,b,c\} and \{a,b,d\}. Deleting b from the five-cycle at a leaves a path on \{c,d,e,f\} with endpoints c,d. Its middle edge is \{e,f\}, so its other two edges are one of the matchings

Deleting a from the link at b gives a path with the same endpoints and middle edge. The two paths must use opposite matchings, as in Figure 1. For example, if the first uses M_1, the triangles \{a,c,e\} and \{a,d,f\} exclude their complementary triples \{b,d,f\} and \{b,c,e\}, forcing the second path to use M_2.

The symmetric difference of the two paths is the four-cycle with edges M_1\cup M_2. Every three of its four vertices retain an edge, so the restrictions of the paths to any such triple differ. A transposition (a\ b) preserving the triangle system would make the two restricted link graphs identical, which is impossible.

Finally, if z is the omitted vertex of a five-element subset, every triple containing z has its complementary triple in that subset. The complementary-triple rule therefore recovers its membership in \mathcal T.□

Proof. Theorem 4.3 says that K is pure of dimension three. A witness line x for the facet S has support exactly S. The restriction of f to \mathbb P(x^\perp)\cong\mathbb{RP}^2 uses only labels in C. It uses all six: otherwise the lower bound in Proposition 4.1, applied in dimension three, would be violated. By Lemma 5.2, this restriction supplies a six-label triangle system contained among the triangles of K on C.

There are no additional triangles on C. Otherwise some complementary triples T,C\setminus T would both be triangles of K. Take witness lines for S,T,C\setminus T. In \mathbb R^4 there is a line perpendicular to all three; their supports together contain all ten labels, leaving that line no admissible label.

If a tetrahedron S' were disjoint from S, its four triples would all be triangles on C. This contradicts Lemma 5.3(2).□

Proof. First observe that if two tetrahedra differ by exchanging labels a,b, the same exchange carries their complementary triangle systems to one another. Their complements share five labels, on which both systems are the triangles of K. After the exchange these triangles remain fixed, so the systems agree everywhere by Lemma 5.3(4).

Suppose three tetrahedra T\cup\{a\}, T\cup\{b\}, T\cup\{c\} contain the same triangle T. Put R=[10]\setminus(T\cup\{a,b,c\}), which has four elements. Transporting the complementary systems successively by

preserves the initial system. The composite fixes R and exchanges b and c, contradicting Lemma 5.3(3). Thus at most two tetrahedra contain any triangle. Positive even multiplicity gives exactly two.□

Proof. Every vertex of an edge link has degree two by Lemma 6.2, so each component is a simple cycle. We show that the link contains no path on five distinct vertices.

Suppose such a path has consecutive vertices v_0,v_1,v_2,v_3,v_4. Let R be the three labels outside D\cup\{v_0,v_1,v_2,v_3,v_4\}. The four successive edges give tetrahedra D\cup\{v_i,v_{i+1}\}, whose complementary systems are transported by

The composite fixes R and sends v_4 to v_2. Hence for every pair \{r,s\}\subset R, the first system contains \{v_4,r,s\} if and only if the last contains \{v_2,r,s\}. The latter triple belongs to both complementary sets, so its membership in either system is simply its membership in K. The first system therefore has identical link graphs at v_4 and v_2 when restricted to R. This contradicts Lemma 5.3(3). Every cycle of length at least five contains such a path, proving the assertion.□

Proof. The partner p(v) exists and is unique by Lemma 6.2, applied to S\setminus\{v\}. It lies outside S, since the only tetrahedron on the vertices of S is S itself. The stated disjointness and lower size bound are immediate.

We first prove that replacing v by a=p(v) leaves the collection of blocks unchanged. Put S'=(S\setminus\{v\})\cup\{a\} and let p' be its partner map. We have p'(a)=v. For w\in S\setminus\{v\}, put D=S\setminus\{v,w\} and b=p(w). The link of D contains the edges

If a=b, these form a three-cycle. Since every link vertex has degree two, that cycle is a whole component, and the other tetrahedron on D\cup\{a\} is D\cup\{a,v\}. Thus p'(w)=v. If a\ne b, the displayed edges form the path a,w,v,b on four distinct vertices. Lemma 6.3 forces it to complete to a four-cycle, with the edge \{a,b\}. Thus p'(w)=b=p(w). The two cases are illustrated in Figure 2. We have proved

The block formerly indexed by a is now indexed by v and has the same elements; all other blocks are unchanged.

The tetrahedron S contains all elements of each block except its indexing partner. The same is true after any such exchange, by block invariance. To change the omitted element of a block from q to u, replace the present element u by its partner q. Doing this once in each block where a change is wanted realizes every possible choice of omissions. These sets have cardinality four, since \sum_y(|B_y|-1)=|S|=4.

Finally, suppose a tetrahedron T met each block in at most one element. In each block omit that element if present, and otherwise omit any element. The resulting tetrahedron lies in the union of the blocks and is disjoint from T, contrary to Lemma 6.1.□

Proof. Assume such a map exists and fix a tetrahedron S of its support complex. Let t be the number of blocks in Lemma 6.4, and let R be the set of labels outside their union. The blocks contain 4+t labels in total, so

If t=4, each block has size two and S meets each in just one element, contradicting the last assertion of Lemma 6.4. Thus t\le3.

There is no triangle entirely in R. By purity, such a triangle would extend to a tetrahedron; its one remaining vertex could not supply two vertices in a block. On the other hand, R is contained in [10]\setminus S, where every four labels contain a triangle by Lemmas 6.1 and 5.3(2). Consequently |R|\le3, and therefore t\ge3.

It follows that t=3, |R|=3, and the block sizes are 3,2,2, as in Figure 3.

The complement of S consists of R and one omitted element from each block. Choose two distinct labels r,s\in R. In the triangle system on this complement, \{r,s\} belongs to two distinct triangles. Their third vertices cannot be in R, because R contains no triangle. They are therefore two distinct omitted block elements. Only one block has size three, so at least one of these triangles is \{r,s,z\} with z in a block B of size two.

Every tetrahedron containing \{r,s,z\} must contain two vertices of some block. As r,s lie outside all blocks, its fourth vertex must be the other element of B. Thus this triangle belongs to at most one tetrahedron, contradicting Lemma 6.2.□

Proof of Theorem 1.1. Lemma 2.1, with k=4, proves compactness and identifies the trace-one affine space with Euclidean \mathbb R^9. It also gives the exact diameter \sqrt2. If ten subsets of strictly smaller diameter covered X, Lemma 2.4 would supply an admissible map \mathbb{RP}^3\to\Delta^9. This contradicts Theorem 6.5.□

Proof. The case d=9 is Theorem 1.1. For d=9+s with s\geq1, translate its projector set to

in the nine-dimensional space of trace-zero symmetric matrices. Every y\in Y has \|y\|_F^2=3/4. In an orthogonal copy of \mathbb R^s, choose vectors v_1,\ldots,v_s with Gram matrix

Such vectors exist because the displayed matrix is positive definite. They satisfy \|v_i\|^2=5/4 and \|v_i-v_j\|^2=2 for i\ne j. Therefore the compact set

has diameter \sqrt2: each new point is at that distance from every point of Y\times\{0\} and from every other new point.

A subset of Z of strictly smaller diameter containing a new point can contain neither a second new point nor a point of Y\times\{0\}. Thus a cover of Z by at most d+1=10+s such subsets would leave at most ten subsets to cover Y\times\{0\}. This contradicts Theorem 1.1, since translation preserves diameter and the number of covering sets.□