56. A matrix certificate for a curved domain
On this page · Permalink · Source line 2179
Let \(E=\mathbb R^r\) and let \(\mathbb S^d\) be the real symmetric \(d\times d\) matrices. Fix an injective linear map
Suppose some \(e\in E\) satisfies \(\mathcal J(e)\succ0\). Define
The map \(\mathcal J\) is an embedding, not a many-to-one decoder of a larger positive coordinate space. The cone is an inverse image of the positive semidefinite cone. This is different from the image cone \(V\mathbb R_+^m\) used in Proposition 39.
Proposition 50: Matrix-embedding probability certificate
The cone \(C_{\mathcal J}\) is closed, pointed and generating, and \(\lambda\) is strictly positive on its nonzero elements. Its normalized base is compact. Suppose a finite-alphabet predictive model has \(z_0\in K_{\mathcal J}\), exact mass conservation and identities
Then every \(B_s\) preserves \(C_{\mathcal J}\), and every finite observation word has a nonnegative, normalized probability. Zero maps may use an empty sum. An invertible change of predictive coordinates transports the certificate by
Read proof · Recorded Lean scope · Visual guide: A curved domain can have a finite witness · Permalink · Source line 2197
Proof
Closedness follows from continuity. If both \(z\) and \(-z\) are in the cone, \(\mathcal J(z)\) is both positive and negative semidefinite, hence zero. Injectivity gives \(z=0\). A sufficiently small neighborhood of \(e\) maps into positive definite matrices, so the cone has an interior point and is generating. A nonzero positive semidefinite matrix has positive trace. The trace-one positive semidefinite matrices form a compact set. The image of \(\mathcal J\) is a closed finite-dimensional subspace and its inverse on that image is continuous, so \(K_{\mathcal J}\) is compact.
Every term \(wKXK^{\mathsf T}\) is positive semidefinite when \(X\) is. The intertwining identity therefore proves invariance. Iteration gives invariance of every branch state, and the mass identity normalizes every finite observation tree. Substituting the transported quantities proves the final statement. \(\square\)
This is a sufficient certificate language. It is not asserted that every invariant cone has this form, or that every positive map on a matrix subspace extends through the displayed congruence representation. The extension requirement is stronger than positivity on the represented states alone.
The base norm of Section 46 is also not automatically the ambient trace norm. One always has
because every positive decomposition in \(E\) gives one in the matrix space. Equality holds when \(\mathcal J\) maps onto all of \(\mathbb S^d\), by the positive and negative spectral parts. For a proper matrix subspace, those spectral parts may leave the subspace. Norm transport therefore needs its own stated hypothesis.