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 (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 \[U_i=\{P\in X_k:\operatorname{dist}_F(P,A_i)<\varepsilon_i\}.\] 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 \[\|P-Q\|_F<d_i+2\varepsilon_i =\frac{d_i+\sqrt2}{2}<\sqrt2.\] 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 (Lee 2013, 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, 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 (4) 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 \[\sum_i E_V(Q)_i=k\int\langle Qu,u\rangle\,d\mu_V(u) =\operatorname{tr}Q.\]

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 \(i\)th integrand is positive at that \(u\), and hence on a nonempty open set of the sphere. Such a set has positive measure. This proves (5) 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 \[B\longmapsto\int_{\mathbb S(V)} \|\pi_Wu\|^2\,\mathbf1_{\{\pi_Wu\ne0,\; \pi_Wu/\|\pi_Wu\|\in B\}}\,d\mu_V(u)\] 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 \[k\int_{\mathbb S(V)}H(Q^{1/2}u)\,d\mu_V(u) =r\int_{\mathbb S(W)}H((Q|_W)^{1/2}w)\,d\mu_W(w),\] 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 \[E_V(Q)=r\int_{\mathbb S^{r-1}}H(TC^{1/2}w)\,d\mu(w).\] 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 6 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 (5) therefore imply that \(E_V(A_+)\) and \(E_V(A_-)\) have disjoint positive coordinates. There is no cancellation in their difference. This proves (8) and, by summing coordinates, (7).

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 (8), 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 \[\Pi_\pm(A)=\frac12\bigl(I\pm A(A^2)^{-1/2}\bigr),\] 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 6 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 (Hatcher 2002, 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 \(i\)th and \(j\)th 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, \[H_n(M,M\setminus q^{-1}(y)) \cong\bigoplus_{x\in q^{-1}(y)}H_n(M,M\setminus\{x\}).\] 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 8. We prove the remaining assertions by examining an arbitrary facet \(I\). Set \[s=|I|,\qquad J=\{1,\ldots,m\}\setminus I,\qquad a=|J|,\qquad h=N_{k-1}.\] 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 8, applied in \(x_0^\perp\), gives \[ a\ge h,\qquad s=m-a\le N_k-N_{k-1}=k. \tag{10}\]

Proof. By Theorem 10, 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 \[M_1=\bigl\{\{c,e\},\{d,f\}\bigr\},\qquad M_2=\bigl\{\{c,f\},\{d,e\}\bigr\}.\] 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 10 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 8, applied in dimension three, would be violated. By Lemma 12, 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 13(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 13(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 \[R\cup\{b,c\} \xrightarrow{(a\ b)}R\cup\{a,c\} \xrightarrow{(b\ c)}R\cup\{a,b\} \xrightarrow{(c\ a)}R\cup\{b,c\}\] preserves the initial system. The composite fixes \(R\) and exchanges \(b\) and \(c\), contradicting Lemma 13(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 15, 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 \[\begin{aligned} R\cup\{v_2,v_3,v_4\} &\xrightarrow{(v_0\ v_2)} R\cup\{v_0,v_3,v_4\} \xrightarrow{(v_1\ v_3)} R\cup\{v_0,v_1,v_4\}\\ &\xrightarrow{(v_2\ v_4)} R\cup\{v_0,v_1,v_2\}. \end{aligned}\] 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 13(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 15, 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 \[\{v,w\},\qquad \{w,a\},\qquad \{v,b\}.\] 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 16 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 \[p'(a)=v,\qquad p'(w)= \begin{cases} v,&p(w)=a,\\ p(w),&p(w)\ne a. \end{cases}\]

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 14. ◻

Proof. Assume such a map exists and fix a tetrahedron \(S\) of its support complex. Let \(t\) be the number of blocks in Lemma 17, and let \(R\) be the set of labels outside their union. The blocks contain \(4+t\) labels in total, so \[1\le t\le4,\qquad |R|=6-t.\] If \(t=4\), each block has size two and \(S\) meets each in just one element, contradicting the last assertion of Lemma 17. 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 14 and 13(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 15. ◻

Proof of Theorem 1. Lemma 2, 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 5 would supply an admissible map \(\mathbb{RP}^3\to\Delta^9\). This contradicts Theorem 18. ◻

Proof. The case \(d=9\) is Theorem 1. For \(d=9+s\) with \(s\geq1\), translate its projector set to \[Y=\{P_x-\tfrac14 I:x\in\mathbb{RP}^3\}\] 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 \[(\langle v_i,v_j\rangle)_{i,j=1}^s =I_s+\tfrac14\mathbf1\mathbf1^{\mathsf T}, \qquad \mathbf1=(1,\ldots,1)^{\mathsf T}.\] 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 \[Z=(Y\times\{0\})\cup\{(0,v_i):1\leq i\leq s\} \subset\mathbb R^9\oplus\mathbb R^s\] 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, since translation preserves diameter and the number of covering sets. ◻