Example. A stable prism line and a partial eigenspace [fdip-005O]

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In the twisted prism, let \(e_u\) and \(e_l\) be the normalized constant vectors on the upper and lower triangles. The group \(\Gamma =C_3\) fixes each, and its trivial isotypic block has multiplicity two. In this orthonormal basis the precision and covariance are \[\begin {aligned} Q_{\mathrm {triv}} &=\begin {pmatrix}5/2&-3/2\\-3/2&5/2\end {pmatrix},\\ Q_{\mathrm {triv}}^{-1} &=\begin {pmatrix}5/8&3/8\\3/8&5/8\end {pmatrix}. \end {aligned}\] Their product is the identity; the determinant of the first is \(4\). The other isotypic block is orthogonal and uncoupled.

Select \(X=\mathbb R e_u\), the line of fields constant on the upper triangle and zero on the lower one. This line is \(\Gamma \)-stable, but \(Qe_u=(5/2)e_u-(3/2)e_l\notin X\). Its compressed covariance is \(5/8\), so \[S_P=\frac 85,\qquad PQ|_X=\frac 52.\] Equivalently, eliminating the lower constant line gives \[\frac 52-\frac {(3/2)^2}{5/2}=\frac 85.\] Thus this subrepresentation induces a different precision. By contrast, the lines spanned by \(e_u+e_l\) and \(e_u-e_l\) have precision eigenvalues \(1\) and \(4\), respectively. Their orthogonal selectors restrict, with retained precisions \(1\) and \(4\). This directly verifies the multiplicity-subspace criterion.

The word eigenvectors in that criterion cannot be replaced by whole eigenspaces. Let the trivial group act on the unit four-cycle. The precision has eigenvalue \(3\) on the two-dimensional space spanned by \[v=(1,0,-1,0),\qquad w=(0,1,0,-1).\] Direct multiplication gives \(Qv=3v\) and \(Qw=3w\). The selector onto \(\mathbb R v\) is proper within this eigenspace and restricts, with \(S_P=3\) in the unit coordinate \(v/\sqrt 2\). Every subspace is stable under the trivial group, but only \(Q\)-invariant ones meet the restriction criterion.