Exact TT ranks and root-sum-square truncation error

Any TT representation with bond size $r_k$ factors the cut unfolding as $A^{⟨k⟩}=L_k{R}_k$: the columns of $L_k$ are the contracted first $k$ cores indexed by the bond ${\alpha }_k$, and the rows of $R_k$ are the contracted remaining cores with that same bond index. This is a matrix product through an inner dimension $r_k$, so $\mathrm{rank}(A^{⟨k⟩})\le r_k$.

For rounding, let $C_{\mathrm{\ell }}$ be the contraction of cores $\mathrm{\ell }+1,\dots ,d$, with row index the bond after core $\mathrm{\ell }$ and column index the suffix $(i_{\mathrm{\ell }+1},\dots ,i_d)$; the empty suffix gives $C_d=[1]$. Put $R_{\mathrm{\ell }}=C_{\mathrm{\ell }}^{\ast }$. Backward induction from $C_d$ shows $C_{\mathrm{\ell }}{C}_{\mathrm{\ell }}^{\ast }=I$: if $G_{\mathrm{\ell }+1}^{(R)}$ is the right matricization of core $G_{\mathrm{\ell }+1}$ and $C_{\mathrm{\ell }+1}{C}_{\mathrm{\ell }+1}^{\ast }=I$, then $C_{\mathrm{\ell }}=G_{\mathrm{\ell }+1}^{(R)}(I_{n_{\mathrm{\ell }+1}}\otimes C_{\mathrm{\ell }+1})$, so $C_{\mathrm{\ell }}{C}_{\mathrm{\ell }}^{\ast }=G_{\mathrm{\ell }+1}^{(R)}(I_{n_{\mathrm{\ell }+1}}\otimes C_{\mathrm{\ell }+1}{C}_{\mathrm{\ell }+1}^{\ast })G_{\mathrm{\ell }+1}^{(R)\ast }=G_{\mathrm{\ell }+1}^{(R)}{G}_{\mathrm{\ell }+1}^{(R)\ast }=I$ by right-isometry. Forward induction similarly gives orthonormal columns for each prefix interface: $L_0=[1]$ and $L_j=(L_{j-1}\otimes I_{n_j})G_j^{(L)}$. Since $L_{j-1}^{\ast }{L}_{j-1}=I$, $(L_{j-1}\otimes I_{n_j}{)}^{\ast }(L_{j-1}\otimes I_{n_j})=I$, so $L_j^{\ast }{L}_j=G_j^{(L)\ast }{G}_j^{(L)}=I$. When the right-to-left sweep processes cut $k$, cores $1,\dots ,k$ still have that left gauge and cores $k+2,\dots ,d$ have the right gauge. The current cut unfolding is $L_k{M}_{k+1}J$, where $M_{k+1}$ is core $G_{k+1}$ reshaped with row index the incoming bond and column index $(i_{k+1},{\alpha }_{k+1})$, and $J=I_{n_{k+1}}\otimes R_{k+1}^{\ast }$. Since $R_{k+1}^{\ast }{R}_{k+1}=I$, $J{J}^{\ast }=I$. The local SVD truncation replaces $M_{k+1}$ by $M_{k+1}{V}_q{V}_q^{\ast }$, where $V_q$ has the retained right singular vectors, so the full unfolding changes from $L_k{M}_{k+1}J$ to $L_k{M}_{k+1}{V}_q{V}_q^{\ast }J=(L_k{M}_{k+1}J)(J^{\ast }{V}_q{V}_q^{\ast }J)$. Since $J{J}^{\ast }=I$ and $V_q^{\ast }{V}_q=I$, $J^{\ast }{V}_q{V}_q^{\ast }J$ is the orthogonal projection onto $T_k=\mathrm{ran}(J^{\ast }{V}_q)$, a subspace of the suffix modes $k+1,\dots ,d$. Thus the resulting tensor is the projection onto $S_k={\mathbb{F}}^{n_1\times \cdots \times n_k}\otimes T_k$. With $E=M_{k+1}(I-V_q{V}_q^{\ast })$, the global residual norm satisfies $\mathrm{\parallel }{L}_{k}EJ{\mathrm{\parallel }}_F^2=\mathrm{tr}(J^{\ast }{E}^{\ast }{L}_k^{\ast }{L}_{k}EJ)=\mathrm{tr}(E^{\ast }EJ{J}^{\ast })=\mathrm{\parallel }E{\mathrm{\parallel }}_F^2={\delta }_k^2$. At the next cut $k-1$, the right interface $R_k$ has orthonormal columns spanning $T_k$. Its local embedding is $I_{n_k}\otimes R_k$, so every newly retained suffix vector lies in $\mathrm{ran}(I_{n_k}\otimes R_k)={\mathbb{F}}^{n_k}\otimes T_k$. Hence $S_{k-1}\subseteq S_k$.

In full-rank TT-SVD, set $Q_0=[1]$ and recursively $Q_k=(Q_{k-1}\otimes I_{n_k})U_k$ using the left singular vectors retained at cut $k$. Since $Q_{k-1}^{\ast }{Q}_{k-1}=I$ and $U_k^{\ast }{U}_k=I$, the columns of every $Q_k$ are orthonormal. Substituting each exact SVD of the current remainder into the preceding factorization gives $A^{⟨k⟩}=Q_k{\mathrm{\Sigma }}_k{V}_k^{\ast }$ for each cut. The retained singular values are all positive, so this factorization has rank $r_k$. By step 1.1 no exact TT can have a smaller bond, proving the full-rank characterization.

For TT-SVD, let $S_0={\mathbb{F}}^{n_1\times \cdots \times n_d}$ and, after truncating cut $k$, let $S_k=\mathrm{ran}(Q_k)\otimes {\mathbb{F}}^{n_{k+1}\times \cdots \times n_d}$. The retained prefix matrices satisfy $\mathrm{ran}(Q_k)\subseteq \mathrm{ran}(Q_{k-1}\otimes I_{n_k})$, hence $S_k\subseteq S_{k-1}$. At step $k$, the current unfolding is $(Q_{k-1}\otimes I_{n_k})M_k$. If $U_q$ contains the retained left singular vectors, the new unfolding is $(Q_{k-1}\otimes I_{n_k})U_q{U}_q^{\ast }{M}_k$, the orthogonal projection of the current tensor onto $S_k$. The discarded matrix $M_k-U_q{U}_q^{\ast }{M}_k$ has squared Frobenius norm ${\delta }_k^2$ by [F3]; left multiplication by the isometry $Q_{k-1}\otimes I_{n_k}$ preserves that norm by direct expansion of the Frobenius inner product, equivalently [F2].

In either algorithm, let $A_0=A$ and let $A_j$ be the approximation after the $j$th truncation in execution order. By steps 3.1 and 1.2, each step is an orthogonal projection onto a retained subspace nested inside the preceding one. Therefore $e_j=A_{j-1}-A_j$ is orthogonal to every later increment, which lies in the retained subspace at step $j$. Telescoping gives $A-\widehat{A}={\sum }_j{e}_j$, an orthogonal sum, and each $\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_F$ is the recorded tail for that cut. Hence $\mathrm{\parallel }A-\widehat{A}{\mathrm{\parallel }}_F^2={\sum }_{k=1}^{d-1}{\delta }_k^2$. If a zero-rank truncation occurs, the output becomes zero and the unperformed tails are zero, so this same argument includes that branch. If no positive singular value is discarded, every ${\delta }_k=0$ and the output is exactly $A$.