subfamilies $\\{W_{k}\\}_{k\in\mathcal{K}}$ and
$\\{U_{\ell}\\}_{\ell\in\mathcal{L}}$ are also transversal to each other.

###### Proof.

Assume that there are two subfamilies $\\{W_{k}\\}_{k\in\mathcal{K}}$ and
$\\{U_{\ell}\\}_{\ell\in\mathcal{L}}$ that are not transversal to each other.
That is,

$\dim\left(\sum_{k\in\mathcal{K}}W_{k}+\sum_{\ell\in\mathcal{L}}U_{\ell}\right)<\sum_{k\in\mathcal{K}}\dim(W_{k})+\sum_{\ell\in\mathcal{L}}\dim(U_{\ell}).$

Then, it follows that

$\displaystyle\dim\left(\sum_{i\in\mathcal{I}}W_{i}+\sum_{j\in\mathcal{J}}U_{j}\right)$
$\displaystyle=\dim\left(\sum_{k\in\mathcal{K}}W_{k}+\sum_{i\in\mathcal{I}\setminus\mathcal{K}}W_{i}+\sum_{\ell\in\mathcal{L}}U_{\ell}+\sum_{j\in\mathcal{J}\setminus\mathcal{L}}U_{j}\right)$
$\displaystyle\leq\dim\left(\sum_{k\in\mathcal{K}}W_{k}+\sum_{\ell\in\mathcal{L}}U_{\ell}\right)+\dim\left(\sum_{i\in\mathcal{I}\setminus\mathcal{K}}W_{i}+\sum_{j\in\mathcal{J}\setminus\mathcal{L}}U_{j}\right)$
$\displaystyle<\sum_{k\in\mathcal{K}}\dim(W_{k})+\sum_{\ell\in\mathcal{L}}\dim(U_{\ell})+\sum_{i\in\mathcal{I}\setminus\mathcal{K}}\dim(W_{i})+\sum_{j\in\mathcal{J}\setminus\mathcal{L}}\dim(U_{\ell})$
$\displaystyle=\sum_{i\in\mathcal{I}}\dim(W_{i})+\sum_{j\in\mathcal{J}}\dim(U_{j}).$

Therefore, $\\{W_{i}\\}_{i\in\mathcal{I}}$ and $\\{U_{j}\\}_{j\in\mathcal{J}}$
are not transversal to each other. ∎

######  Corollary B.3.

Let $\\{W_{i}\\}_{i\in\mathcal{I}}$ be a transverse family. Then, for any
$\mathcal{J}\subseteq\mathcal{I}$, the subfamily
$\\{W_{j}\\}_{j\in\mathcal{J}}$ is also a transverse family.

###### Proof.

Apply Lemma B.2 to $\\{W_{i}\\}_{i\in\mathcal{I}}$ and $\\{U:=\\{0\\}\\}$ ∎

######  Lemma B.4.

Let $\mathcal{I}$ be a finite set and
$\mathsf{M}:\mathcal{I}\to\mathsf{Gr}(V)$ be any function such that
$\\{\mathsf{M}(i)\\}_{i\in\mathcal{I}}$ is a transverse family. Let
$\mathcal{J}$ be any finite set and $h:\mathcal{I}\to\mathcal{J}$ be any
function. Then,

$\sum_{j\in\mathcal{J}}\dim(h_{\sharp}\mathsf{M}(j))=\sum_{i\in\mathcal{I}}\dim(\mathsf{M}(i)).$

In particular, the family $\\{h_{\sharp}\mathsf{M}(i)\\}_{j\in J}$ is a
transverse family.

###### Proof.

The claim follows from the following calculation:

$\displaystyle\sum_{j\in\mathcal{J}}\dim(h_{\sharp}\mathsf{M}(j))=$
$\displaystyle\sum_{j\in\mathcal{J}}\dim\left(\sum_{i\in
h^{-1}(j)}\mathsf{M}(i)\right)$
$\displaystyle=\sum_{j\in\mathcal{J}}\sum_{i\in h^{-1}(j)}\dim(\mathsf{M}(i))$
by Corollary B.3 $\displaystyle=\sum_{i\in\mathcal{I}}\dim(\mathsf{M}(i))$
$\displaystyle=\dim\left(\sum_{i\in\mathcal{I}}\mathsf{M}(i)\right)=\dim\left(\sum_{j\in\mathcal{J}}h_{\sharp}\mathsf{M}(j)\right).$

∎

###### Proof of Proposition 4.14.

For $i=1,2,3$, let $\mathsf{M}_{i}:\bar{P}_{i}^{\times}\to\mathsf{Gr}(V)$ be
Grassmannian persistence diagrams. Let $(f,\zeta_{P_{2}})$ be a morphism from
$\mathsf{M}_{1}$ to $\mathsf{M}_{2}$ and let $(g,\zeta_{P_{3}})$ be a morphism
from $\mathsf{M}_{2}$ to $\mathsf{M}_{3}$. Thus, we have that
$\left(\overline{f_{\bullet}}\right)_{\sharp}\mathsf{M}_{1}\simeq_{\mathsf{M\ddot{o}b}}(\mathsf{M}_{2}+\zeta_{P_{2}})$
and
$\left(\overline{g_{\bullet}}\right)_{\sharp}\mathsf{M}_{2}\simeq_{\mathsf{M\ddot{o}b}}(\mathsf{M}_{3}+\zeta_{P_{3}})$.
Then, it follows that

$\displaystyle\left(\overline{(g\circ
f)_{\bullet}}\right)_{\sharp}\mathsf{M}_{1}$
$\displaystyle=\left(\overline{g_{\bullet}\circ
f_{\bullet}}\right)_{\sharp}\mathsf{M}_{1}$
$\displaystyle=\left(\overline{g_{\bullet}}\circ\overline{f_{\bullet}}\right)_{\sharp}\mathsf{M}_{1}$
$\displaystyle=\left(\overline{g_{\bullet}}\right)_{\sharp}\left(\left(\overline{f_{\bullet}}\right)_{\sharp}\mathsf{M}_{1}\right)$
$\displaystyle\simeq_{\mathsf{M\ddot{o}b}}\left(\overline{g_{\bullet}}\right)_{\sharp}\left(\mathsf{M}_{2}+\zeta_{P_{2}}\right)$
by Lemma B.1
$\displaystyle=\left(\overline{g_{\bullet}}\right)_{\sharp}\left(\mathsf{M}_{2}\right)+\left(\overline{g_{\bullet}}\right)_{\sharp}\left(\zeta_{P_{2}}\right)$
$\displaystyle\simeq_{\mathsf{M\ddot{o}b}}\mathsf{M}_{3}+\zeta_{P_{3}}+\left(\overline{g_{\bullet}}\right)_{\sharp}\left(\zeta_{P_{2}}\right).$

Observe that
$\zeta_{{P_{3}}^{\prime}}:=\zeta_{P_{3}}+\left(\overline{g_{\bullet}}\right)_{\sharp}(\zeta_{P_{2}})$
is supported on $\mathsf{diag}(P_{3})$ and the families
$\left\\{\zeta_{P_{3}}(J)\right\\}_{J\in\bar{P}_{3}^{\times}}$ and
$\\{(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}}(J))\\}_{J\in\bar{P}_{3}^{\times}}$
are transversal to each other. The latter can be seen from the following
argument. The families $\\{\zeta_{P_{2}}(I)\\}_{I\in\bar{P}_{2}^{\times}}$ and
$\\{\mathsf{M}_{2}(I)\\}_{I\in\bar{P}_{2}^{\times}}$ are transversal to each
other. Thus,
$\\{(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}}(J))\\}_{J\in\bar{P}_{3}^{\times}}$
and $\\{\mathsf{M}_{2}(I)\\}_{I\in\bar{P}_{2}^{\times}}$ are transversal to
each other. Moreover, we have that
$\sum_{I\in\bar{P}_{2}^{\times}}\mathsf{M}_{2}(I)=\sum_{J\in\bar{P}_{3}^{\times}}\mathsf{M}_{3}(J)+\zeta_{P_{3}}(J)$
and
$\\{\mathsf{M}_{3}(J)\\}_{J\in\bar{P}_{3}^{\times}}\cup\\{\zeta_{P_{3}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$
is a transversal family. Hence,
$\\{(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}}(J))\\}_{J\in\bar{P}_{3}^{\times}}$
and
$\\{\mathsf{M}_{3}(J)\\}_{J\in\bar{P}_{3}^{\times}}\cup\\{\zeta_{P_{3}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$
are transversal to each other. Thus,
$\\{(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}}(J))\\}_{J\in\bar{P}_{3}^{\times}}$
and $\\{\zeta_{P_{3}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$ are transversal to
each other.

It remains to show that $\\{\mathsf{M}_{3}(J)\\}_{J\in\bar{P}_{3}^{\times}}$
and $\\{\zeta_{{P_{3}}^{\prime}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$ are
transversal to each other. That is, we need to show that

$\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}\mathsf{M}_{3}(J)+\sum_{J\in\bar{P}_{3}^{\times}}\zeta_{{P_{3}}^{\prime}}(J)\right)=\sum_{J\in\bar{P}_{3}^{\times}}\dim(\mathsf{M}_{3}(J))+\sum_{J\in\bar{P}_{3}^{\times}}\dim(\zeta_{{P_{3}}^{\prime}}(J)).$

We have that LHS is equal to

$\displaystyle=\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}\mathsf{M}_{3}(J)+\sum_{J\in\bar{P}_{3}^{\times}}\zeta_{{P_{3}}^{\prime}}(J)\right)$
$\displaystyle=\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}\mathsf{M}_{3}(J)+\sum_{J\in\bar{P}_{3}^{\times}}\zeta_{{P_{3}}}(J)+\sum_{J\in\bar{P}_{3}^{\times}}(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J)\right)$
$\displaystyle=\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}\big{(}\mathsf{M}_{3}(J)+\zeta_{{P_{3}}}(J)\big{)}+\sum_{J\in\bar{P}_{3}^{\times}}(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J)\right)$
$\displaystyle=\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}(\overline{g_{\bullet}})_{\sharp}(\mathsf{M}_{2})(J)+\sum_{J\in\bar{P}_{3}^{\times}}(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J)\right)$
as
$(\overline{g_{\bullet}})_{\sharp}\mathsf{M}_{2}\simeq_{\mathsf{M\ddot{o}b}}(\mathsf{M}_{3}+\zeta_{P_{3}})$
$\displaystyle=\dim\left(\sum_{I\in\bar{P}_{2}^{\times}}\mathsf{M}_{2}(I)+\sum_{I\in\bar{P}_{2}^{\times}}\zeta_{P_{2}}(I)\right)$
$\displaystyle=\sum_{I\in\bar{P}_{2}^{\times}}\dim(\mathsf{M}_{2}(I))+\sum_{I\in\bar{P}_{2}^{\times}}\dim(\zeta_{P_{2}}(I))$
$\displaystyle=\dim\left(\sum_{I\in\bar{P}_{2}^{\times}}\mathsf{M}_{2}(I)\right)+\sum_{J\in\bar{P}_{3}^{\times}}\dim((\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J))$
by Lemma B.4
$\displaystyle=\dim\left(\sum_{J\in\bar{P}_{3}^{\times}}\mathsf{M}_{3}(J)+\sum_{J\in\bar{P}_{3}^{\times}}\zeta_{P_{3}}(J)\right)+\sum_{J\in\bar{P}_{3}^{\times}}\dim((\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J))$
$\displaystyle=\sum_{J\in\bar{P}_{3}^{\times}}\dim(\mathsf{M}_{3}(J))+\sum_{J\in\bar{P}_{3}^{\times}}\dim(\zeta_{P_{3}}(J))+\sum_{J\in\bar{P}_{3}^{\times}}\dim((\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J))$
$\displaystyle=\sum_{J\in\bar{P}_{3}^{\times}}\dim(\mathsf{M}_{3}(J))+\sum_{J\in\bar{P}_{3}^{\times}}\dim(\zeta_{{P_{3}}^{\prime}}(J)).$

The last equality follows from Lemma B.2 as we have that the families
$\\{\zeta_{P_{3}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$ and
$\\{(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}}(J)\\}_{J\in\bar{P}_{3}^{\times}}$
are transversal to each other, therefore,

$\dim\left(\zeta_{P_{3}}(J)\right)+\dim\left((\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J)\right)=\dim(\zeta_{P_{3}}(J)+(\overline{g_{\bullet}})_{\sharp}(\zeta_{P_{2}})(J))=\dim(\zeta_{{P_{3}}^{\prime}}(J))$

for all $J\in\bar{P}_{3}^{\times}$. ∎

######  Lemma B.5.

If $W_{1},W_{2}\subseteq V$, are subspaces of an inner product space $V$, then

$W_{1}\ominus W_{2}=W_{1}\ominus\operatorname{proj}_{W_{1}}(W_{2}).$

###### Proof.

Let $u\in W_{1}\cap W_{2}^{\perp}$ and $w_{2}\in W_{2}$. Then,

$\displaystyle 0=$ $\displaystyle\langle u,w_{2}\rangle$ $\displaystyle=$
$\displaystyle\langle
u,\operatorname{proj}_{W_{1}}(w_{2})+(w_{2}-\operatorname{proj}_{W_{1}}(w_{2}))\rangle$
$\displaystyle=$ $\displaystyle\langle
u,\operatorname{proj}_{W_{1}}(w_{2})\rangle+\langle
u,(w_{2}-\operatorname{proj}_{W_{1}}(w_{2}))\rangle$ $\displaystyle=$
$\displaystyle\langle u,\operatorname{proj}_{W_{1}}(w_{2})\rangle+0.$

Thus, $\langle u,\operatorname{proj}_{W_{1}}(w_{2})\rangle=0$. Therefore,
$u\in
W_{1}\cap(\operatorname{proj}_{W_{1}}(W_{2}))^{\perp}=W_{1}\ominus\operatorname{proj}_{W_{1}}(W_{2}).$
Let $s\in W_{1}\cap(\operatorname{proj}_{W_{1}}(W_{2}))^{\perp}$ and let
$w_{2}\in W_{2}$. Then,

$\displaystyle\langle s,w_{2}\rangle=$ $\displaystyle\langle
s,\operatorname{proj}_{W_{1}}(w_{2})+(w_{2}-\operatorname{proj}_{W_{1}}(w_{2}))\rangle$
$\displaystyle=$ $\displaystyle\langle
s,\operatorname{proj}_{W_{1}}(w_{2})\rangle+\langle
s,(w_{2}-\operatorname{proj}_{W_{1}}(w_{2}))\rangle$ $\displaystyle=$
$\displaystyle 0+0=0$

Thus, $s\in W_{1}\cap W_{2}^{\perp}=W_{1}\ominus W_{2}$. Therefore,
$W_{1}\ominus W_{2}=W_{1}\ominus\operatorname{proj}_{W_{1}}(W_{2})$. ∎

######  Lemma B.6.

Let $V$ be a finite-dimensional inner product space. Let $B,C\subseteq V$ be
two linear subspaces. Then,

$B^{\perp}\cap(C\cap(B\cap C)^{\perp})^{\perp}=B^{\perp}\cap C^{\perp}.$

###### Proof.

The containment $B^{\perp}\cap(C\cap(B\cap C)^{\perp})^{\perp}\supseteq
B^{\perp}\cap C^{\perp}$ is clear because $(C\cap(B\cap
C)^{\perp})^{\perp}\supseteq C^{\perp}$ as $C\cap(B\cap C)^{\perp}\subseteq
C$. To see the other containment $B^{\perp}\cap(C\cap(B\cap
C)^{\perp})^{\perp}\subseteq B^{\perp}\cap C^{\perp}$, let $x\in
B^{\perp}\cap(C\cap(B\cap C)^{\perp})^{\perp}$. Then, $x\in B^{\perp}$ and
$x\in(C\cap(B\cap C)^{\perp})^{\perp}=C^{\perp}+(B\cap C)$. Thus, we can write
$x=w+y$ where $w\in C^{\perp}$ and $y\in B\cap C$. Then, we have

$\displaystyle 0$ $\displaystyle=\langle x,y\rangle$ as $x\in B^{\perp}$ and
$y\in B$ $\displaystyle=\langle w+y,y\rangle$ $\displaystyle=\langle
w,y\rangle+\langle y,y\rangle$ $\displaystyle=0+\langle y,y\rangle$ as $w\in
C^{\perp}$ and $y\in C$ $\displaystyle=\langle y,y\rangle.$

Thus, $y=0$ and $x=w\in C^{\perp}$. Therefore, $x\in B^{\perp}\cap C^{\perp}$.
Hence, $B^{\perp}\cap(C\cap(B\cap C)^{\perp})^{\perp}\subseteq B^{\perp}\cap
C^{\perp}$. ∎

###### Proof of Proposition 4.23.

Unraveling the definition of $\ominus$, we see that the desired equality is
equivalent to

$(A\cap B^{\perp})\cap(C\cap(B\cap C)^{\perp})^{\perp}=A\cap(B+C)^{\perp}.$

By Lemma B.6, we have that

$B^{\perp}\cap(C\cap(B\cap C)^{\perp})^{\perp}=(B+C)^{\perp}$

as $B^{\perp}\cap C^{\perp}=(B+C)^{\perp}$. Intersecting both sides with $A$
provides the desired equality.

∎

##  Appendix C Edit distance stability of barcodes and orthogonal inversions

In this section, we revisit some notions of the functorial pipeline for
persistent homology proposed by McCleary and Patel in [21] in order to
establish the connections between their work and ours.

#### Non-negative Integral Functions (Persistence Diagrams / Barcodes).

Let $P$ and $Q$ be two finite linearly ordered metric posets. Let
$\alpha:P\to\mathbb{Z}_{\geq 0}$ and $\beta:Q\to\mathbb{Z}_{\geq 0}$ be two
(not necessarily order-preserving) functions (also called persistence diagrams
or barcodes). As the target of $\alpha$ and $\beta$ is the set of non-negative
integers, $\mathbb{Z}_{\geq 0}$, we call them non-negative integral functions.
A  _charge-preserving morphism_ from $\alpha$ to $\beta$ is any Galois
connection $f_{\bullet}:P\leftrightarrows Q:f^{\bullet}$ such that

$\beta(J)=\sum_{I\in\left(\overline{f_{\bullet}}\right)^{-1}(J)}\alpha(I)$

for every $J\in\mathsf{Int}(Q)\setminus\mathsf{diag}(Q)$.

Let $\mathsf{Fnc}$ denote the category whose objects are non-negative integral
functions and whose morphisms are charge-preserving morphisms.

######  Definition C.1 (Cost of a morphism in $\mathsf{Fnc}$).

The _cost_ of a morphism $f=(f_{\bullet},f^{\bullet})$ in $\mathsf{Fnc}$,
denoted $\mathsf{cost}_{\mathsf{Fnc})}(f)$, is defined to be
$\mathsf{cost}_{\mathsf{Fnc}}(f):=||f_{\bullet}||$, the distortion of the left
adjoint $f_{\bullet}$.

Let $\mathsf{F}:P\to\mathsf{SubCx}(K)$ be a filtration. The $q$-th birth-death
function $\mathsf{F}$, denoted by
$\dim\left(\mathsf{ZB}_{q}^{\mathsf{F}}\right):\bar{P}^{\times}\to\mathbb{Z}$,
is defined to be the dimension of $q$-th the birth-death space of $\mathsf{F}$
for every interval.

$\displaystyle\dim\left(\mathsf{ZB}_{q}^{\mathsf{F}}\right):\bar{P}^{\times}$
$\displaystyle\to\mathbb{Z}$ $\displaystyle[i,j]$
$\displaystyle\mapsto\dim\left(\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])\right).$

######  Proposition C.2 ([21, Proposition 5.8, Proposition 6.6]).

Let $\mathsf{F}:P\to\mathsf{SubCx}(K)$ be a filtration. Then, for any degree
$q\geq 0$, the assignment

$\mathsf{F}\mapsto\partial_{\bar{P}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{F}})$

is a functor from $\mathsf{Fil}(K)$ to $\mathsf{Fnc}$.

As a result of this functoriality, Proposition C.2, and the fact that
$\partial_{\bar{P}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{F}})$ is precisely
the degree-$q$ persistence diagram of $\mathsf{F}$, the edit distance
stability of persistence diagrams follows.

######  Theorem C.3 ([21, Theorem 8.4 (Stability)]).

Let $\mathsf{F}:P\to\mathsf{SubCx}(K)$ and $\mathsf{G}:Q\to\mathsf{SubCx}(K)$
be two filtrations of a finite simplicial complex $K$ indexed by finite
linearly ordered metric posets and let
$\partial_{\bar{P}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{F}}):\bar{P}^{\times}\to\mathbb{Z}_{\geq
0}$ and
$\partial_{\bar{Q}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{G}}):\bar{Q}^{\times}\to\mathbb{Z}_{\geq
0}$ be their respective degree-$q$ persistence diagrams. Then,

$d_{\mathsf{Fnc}}^{E}(\partial_{\bar{P}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{F}}),\partial_{\bar{Q}^{\times}}\dim(\mathsf{ZB}_{q}^{\mathsf{G}}))\leq
d_{\mathsf{Fil}(K)}^{E}(\mathsf{F},\mathsf{G}).$

Note that our stability theorem, Theorem 5.6, which states that

$d_{\mathsf{GrPD}(C_{q}^{K})}^{E}(\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}},\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{G}})\leq
d_{\mathsf{Fil}(K)}^{E}(\mathsf{F},\mathsf{G})$

is analogous to Theorem C.3. As a matter of fact, Theorem 5.6 improves the
stability result of McCleary and Patel, Theorem C.3, in the following sense.

######  Theorem C.4.

Let $\mathsf{F}:P\to\mathsf{SubCx}(K)$ and $\mathsf{G}:Q\to\mathsf{SubCx}(K)$
be two filtrations of a finite simplicial complex $K$ indexed by finite
linearly ordered metric posets. Then, for any degree $q\geq 0$, we have

$d_{\mathsf{Fnc}}^{E}\left(\partial_{\bar{P}^{\times}}\dim\left(\mathsf{ZB}_{q}^{\mathsf{F}}\right),\partial_{\bar{Q}^{\times}}\dim\left(\mathsf{ZB}_{q}^{\mathsf{G}}\right)\right)\leq
d_{\mathsf{GrPD}(C_{q}^{K})}^{E}\left(\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}},\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{G}}\right)\leq
d_{\mathsf{Fil}(K)}^{E}\left(\mathsf{F},\mathsf{G}\right).$

In other words, the edit distance between the degree-$q$ Grassmannian
persistence diagrams mediates between the edit distance between classical
degree-$q$ persistence diagrams and the edit distance between filtrations. We
will need the following lemmas and proposition for proving Theorem C.4.

######  Lemma C.5.

Let $\mathcal{I}$ be a finite set and let the two families
$\\{W(I)\\}_{I\in\mathcal{I}}$ and $\\{U(I)\\}_{I\in\mathcal{I}}$ be
transversal to each other. Then, the family
$\\{W(I)+U(I)\\}_{I\in\mathcal{I}}$ is a transverse family.

###### Proof.

We have

$\displaystyle\dim\left(\sum_{i\in\mathcal{I}}(W(i)+U(i))\right)$
$\displaystyle=\dim\left(\sum_{i\in\mathcal{I}}W(i)+\sum_{i\in\mathcal{I}}U(i)\right)$
$\displaystyle=\sum_{i\in\mathcal{I}}\dim(W(i))+\sum_{i\in\mathcal{I}}\dim(U(i))$
$\displaystyle=\sum_{i\in\mathcal{I}}(\dim(W(i))+\dim(U(i)))$
$\displaystyle=\sum_{i\in\mathcal{I}}\dim(W(i)+U(i)).$

The last equality follows from the fact that
$\dim(W(i)+U(i))=\dim(W(i))+\dim(U(i))$, which can be derived from Lemma B.2.
∎

######  Lemma C.6.

Let $R$ and $S$ be two finite posets. Let $\mathsf{M}_{1}:R\to\mathsf{Gr}(V)$
and $\mathsf{M}_{2}:S\to\mathsf{Gr}(V)$ be two functions such that
$\\{\mathsf{M}_{1}(r)\\}_{r\in R}$ and $\\{\mathsf{M}_{2}(s)\\}_{s\in S}$ are
transversal families. Assume that $h:R\to S$ is an order-preserving map such
that $h_{\sharp}\mathsf{M}_{1}\simeq_{\mathsf{M\ddot{o}b}}\mathsf{M}_{2}$.
Then, for every $s\in S$, it holds that

$\dim(\mathsf{M}_{2}(s))=\sum_{r\in h^{-1}(s)}\dim(\mathsf{M}_{1}(r)).$

###### Proof.

We will proceed by induction on $s\in S$. For the base cases, let $0_{s}$ be a
minimal element of $S$ (note that there could be more than one minimal element
of $S$). As we have that
$h_{\sharp}\mathsf{M}_{1}\simeq_{\mathsf{M\ddot{o}b}}\mathsf{M}_{2}$, it
follows that

$\mathsf{M}_{2}(0_{s})=\sum_{r\in h^{-1}(0_{s})}\mathsf{M}_{1}(r).$

Considering the dimension of both sides, we obtain

$\displaystyle\dim(\mathsf{M}_{2}(0_{s}))$ $\displaystyle=\dim\left(\sum_{r\in
h^{-1}(0_{s})}\mathsf{M}_{1}(r)\right)$ $\displaystyle=\sum_{r\in
h^{-1}(0_{s})}\dim(\mathsf{M}_{1}(r)),$

where the last equality follows from Corollary B.3. Now, let $s\in S$ be fixed
and assume that for every $q\in S$ such that $q<s$, it holds that

$\dim(\mathsf{M}_{2}(q))=\sum_{r\in h^{-1}(q)}\dim(\mathsf{M}_{1}(r)).$ (10)

Using the Möbius equivalence
$\mathsf{M}_{2}\simeq_{\mathsf{M\ddot{o}b}}h_{\sharp}\mathsf{M}_{1}$, we see
that

$\displaystyle\dim\left(\sum_{q\leq s}\mathsf{M}_{2}(q)\right)$
$\displaystyle=\dim\left(\sum_{q\leq s}h_{\sharp}\mathsf{M}_{1}(q)\right)$
$\displaystyle=\dim\left(\sum_{\begin{subarray}{c}r\in R\\\ h(r)\leq
s\end{subarray}}\mathsf{M}_{1}(r)\right)$
$\displaystyle=\sum_{\begin{subarray}{c}r\in R\\\ h(r)\leq
s\end{subarray}}\dim(\mathsf{M}_{1}(r))$ by Corollary B.3
$\displaystyle=\sum_{\begin{subarray}{c}r\in R\\\
h(r)<s\end{subarray}}\dim(\mathsf{M}_{1}(r))+\sum_{r\in
h^{-1}(s)}\dim(\mathsf{M}_{1}(s))$
$\displaystyle=\sum_{q<s}\dim(\mathsf{M}_{2}(q))+\sum_{r\in
h^{-1}(s)}\dim(\mathsf{M}_{1}(s)),$

where the last equality follows from our induction hypothesis Eq. 10. On the
other hand, by Corollary B.3, we have that

$\displaystyle\dim\left(\sum_{q\leq s}\mathsf{M}_{2}(q)\right)$
$\displaystyle=\sum_{q\leq s}\dim(\mathsf{M}_{2}(q))$
$\displaystyle=\sum_{q<s}\dim(\mathsf{M}_{2}(q))+\dim(\mathsf{M}_{2}(s)).$

Hence, we conclude that

$\dim(\mathsf{M}_{2}(s))=\sum_{r\in h^{-1}(s)}\dim(\mathsf{M}_{1}(r))$

for every $s\in S$. ∎

######  Proposition C.7.

Let $V$ be a vector space and let
$\mathsf{M}:\bar{P}^{\times}\to\mathsf{Gr}(V)$ be an object in
$\mathsf{GrPD}(V)$. The assignment

$\mathsf{M}\mapsto\dim(\mathsf{M})$

is a functor from $\mathsf{GrPD}(V)$ to $\mathsf{Fnc}$, where
$\dim(\mathsf{M}):\bar{P}^{\times}\to\mathbb{Z}_{\geq 0}$ is defined by

$\dim(\mathsf{M})(I):=\dim(\mathsf{M}(I))$

for $I\in\bar{P}^{\times}$.

###### Proof.

Let $\mathsf{M}:\bar{P}^{\times}\to\mathsf{Gr}(V)$ and
$\mathsf{N}:\bar{Q}^{\times}\to\mathsf{Gr}(V)$ be two objects in
$\mathsf{GrPD}(V)$ and let $f_{\bullet}:P\leftrightarrows Q:f^{\bullet}$ be a
morphism from $\mathsf{M}$ to $\mathsf{N}$. This means that there is
$\zeta_{Q}:\bar{Q}^{\times}\to\mathsf{Gr}(V)$ supported on $\mathsf{diag}(Q)$
such that the families $\\{\mathsf{N}(J)\\}_{J\in\bar{Q}^{\times}}$ and
$\\{\zeta_{Q}(J)\\}_{J\in\bar{Q}^{\times}}$ are transversal to each other and

$\left(\overline{f_{\bullet}}\right)_{\sharp}\mathsf{M}\simeq_{\mathsf{M\ddot{o}b}}\mathsf{N}+\zeta_{Q}.$

Let $\tilde{\mathsf{N}}(J):=\mathsf{N}(J)+\zeta_{Q}(J)$ for every
$J\in\bar{Q}^{\times}$. Then, by Lemma C.5, we have that
$\\{\tilde{\mathsf{N}}(J)\\}_{J\in\bar{Q}^{\times}}$ is a transverse family.
Moreover, we have that
$\left(\overline{f_{\bullet}}\right)_{\sharp}\mathsf{M}\simeq_{\mathsf{M\ddot{o}b}}\tilde{\mathsf{N}}$.
Hence, by Lemma C.6, we conclude that

$\dim\left(\tilde{\mathsf{N}}(J)\right)=\sum_{I\in\left(\overline{f_{\bullet}}\right)^{-1}(J)}\dim(\mathsf{M}(I)).$

In particular, for every $J\in\mathsf{Int}(Q)\setminus\mathsf{diag}(Q)$, we
have

$\dim\left(\mathsf{N}(J)\right)=\sum_{I\in\left(\overline{f_{\bullet}}\right)^{-1}(J)}\dim(\mathsf{M}(I)).$

Thus, the Galois connection $f_{\bullet}:P\leftrightarrows Q:f^{\bullet}$ is a
morphism from $\dim(\mathsf{M})$ to $\dim(\mathsf{N})$. ∎

###### Proof of Theorem C.4.

By Proposition C.7, every path in $\mathsf{GrPD}(C_{q}^{K})$ induces a path in
$\mathsf{Fnc}$ with the same cost. Moreover, as already discussed in Section
5, we have

$\dim\left(\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}\right)=\partial_{\bar{P}^{\times}}\dim\left(\mathsf{ZB}_{q}^{\mathsf{F}}\right).$

Thus, the result follows. ∎

Figure 12: Two filtrations $\mathsf{F}$ and $\mathsf{G}$, their treegrams and
degree-$0$ persistence diagrams; see Example C.8

######  Example C.8.

Consider the filtrations $\mathsf{F}$ and $\mathsf{G}$ depicted in Fig. 12.
Their degree-$0$ persistence diagrams are the same, thus

$d_{\mathsf{Fnc}}^{E}\left(\partial_{\bar{P}^{\times}}\dim\left(\mathsf{ZB}_{0}^{\mathsf{F}}\right),\partial_{\bar{Q}^{\times}}\dim\left(\mathsf{ZB}_{0}^{\mathsf{G}}\right)\right)=0.$

On the other hand, $\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}$ and
$\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{G}}$ distinguish the two
filtrations as we have that

$d_{\mathsf{GrPD}(C_{0}^{K})}^{E}\left(\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}},\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{G}}\right)>0.$

Indeed, one can see that
$d_{\mathsf{GrPD}(C_{0}^{K})}^{E}\left(\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}},\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{G}}\right)>0$
by using the equivalence of treegrams and degree $0$ Grassmannian persistence
diagrams, Theorem 5.13. Note that the two filtrations $\mathsf{F}$ and
$\mathsf{G}$ yield two different treegrams, therefore
$\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}$ and
$\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{G}}$ are two different
Grassmannian persistence diagrams.

##  Appendix D An Algorithm for Reconstructing the Degree-$0$ Grassmannian
Persistence Diagram from the Treegram

Let $\mathsf{F}:=\\{K_{i}\\}_{i=1}^{n}$ be a filtration of a connected finite
simplicial complex $K$ and let $T_{\mathsf{F}}$ be the treegram of
$\mathsf{F}$. In this section, we describe an algorithm for reconstructing
$\mathsf{OI}_{\times}\mathsf{ZB}_{0}$ from the treegram $T_{\mathsf{F}}$.

Using the treegram $T_{\mathsf{F}}$, we first form a subspace
$\mathcal{S}([b,d])\subseteq C_{0}^{K}$ for every $1\leq b\leq d\leq n$. Then,
we show that
$\mathcal{S}([b,d])=\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}([b,d])$
for all $b\leq d$ (Proposition D.2). The construction of $\mathcal{S}([b,d])$
involves multiple steps. We proceed through the following steps:

  1. 1.

Fix $1<d\leq n$. (to be treated as a death time)

  2. 2.

Fix a block of the sub-partition $T_{\mathsf{F}}(d)$, say $B_{i}$.

  3. 3.

Detect the blocks of $T_{\mathsf{F}}(d-1)$ that merge into $B_{i}$ at time
$d$, and define a notion of birth times for these blocks, say
$b_{i,1},\ldots,b_{i,m_{i}}$.

  4. 4.

For each birth time $b_{i,j}$ form a subspace $S_{i}([b_{i,j},d])\subseteq
C_{0}^{K}$.

  5. 5.

Repeat step $3$ and step $4$ for every block $B_{i}$ of $T_{\mathsf{F}}(d)$
and for every $1<d\leq n$.

  6. 6.

Form subspaces $\mathcal{S}([b,d])$ by appropriately organizing
$S_{i}([b_{i,j},d])$s.

Step $1$: Let $1<d\leq n$ be fixed.

Step $2$: Let $B_{1},\ldots,B_{N}$ be the blocks of $T_{\mathsf{F}}(d)$. Fix
$1\leq i\leq N$.

Step $3$: Assume that there are block $B_{i,1},\ldots,B_{i,m_{i}}$ at time
$t=d-1$ that merge into $B_{i}$ at time $d$. That is,
$\cup_{j=1}^{m_{i}}B_{i,j}\subseteq B_{i}$. Notice that $B_{i}$ might be
strictly larger than the union since there might be _ephemeral_ points, i.e.
points with the same birth and death time. We let
$\\{v_{i,1},\ldots,v_{i,l_{i}}\\}=B_{i}\setminus(\cup_{j=1}^{m_{i}}B_{i,j})$
denote all such ephemeral points. Recall that we defined the birth time of a
point $x\in X$ in a treegram $T_{X}$ as

$b_{x}:=\min\\{t\in\mathbb{R}\mid x\in X_{t}\\}.$

Let $V:=V(K)$ be the vertex set of $K$. For a non-empty subset $Y\subseteq V$,
we define

$\displaystyle b(Y)$ $\displaystyle:=\min\\{b_{y}\mid y\in Y\\},$
$\displaystyle c(Y)$ $\displaystyle:=\frac{1}{|Y|}\sum_{y\in Y}y\;\;\in
C_{0}^{K}.$

Let $b_{i,j}:=b(B_{i,j})$. Without loss of generality, we may assume that

$b_{i,1}=b_{i,2}=\cdots=b_{i,k_{i}}<b_{i,k_{i}+1}\leq\cdots\leq b_{i,m_{i}}.$

Step $4$: For $1\leq j\leq m_{i}$ we define

$\displaystyle R_{i,j}$ $\displaystyle:=\\{x\in B_{i,j}\mid b_{x}=b_{i,j}\\}$
$\displaystyle c_{i,j}$ $\displaystyle:=c(R_{i,j}).$

For any $0$-chain $c\in C_{0}^{K}$, let $\operatorname{span}\\{c\\}\subseteq
C_{0}^{K}$ denote the subspace generated by $c$. Now, we define

$S_{i}([b_{i,k_{i}},d]):=\sum_{l=2}^{k_{i}}\operatorname{span}\\{c_{i,l}-c_{i,1}\\}\;\;\subseteq
C_{0}^{K}.$

When $k_{i}=1$, then the sum above is an empty sum. In that case, we let
$S_{i}([b_{i,j},d]):=\\{0\\}$. Notice also that
$\dim(S_{i}([b_{i,k_{i}},d]))=k_{i}-1$ , which is the number of connected
components that are born at $b_{k_{i}}$ and dead at $d$ by merging together
into $B_{i}$.

For $k_{i}+1\leq j\leq m_{i}$, we define

$\displaystyle{R^{\prime}}_{i,j}$ $\displaystyle:=\\{x\in B_{i}\mid\exists
B_{i,j^{\prime}}\ni x\text{ such that }b_{x}\leq b_{i,j}\text{ and
}b_{i,j^{\prime}}<b_{i,j}\\},$ $\displaystyle{c^{\prime}}_{i,j}$
$\displaystyle:=c({R^{\prime}}_{i,j}),$ $\displaystyle S_{i}([b_{i,j},d]))$
$\displaystyle:=\operatorname{span}\\{c_{i,j}-{c^{\prime}}_{i,j}\\}\;\;\subseteq
C_{0}^{K}.$

Step $5$: Repeat steps $3$ and $4$ for every block of $T_{\mathsf{F}}(d)$ and
for every $1<d\leq n$.

Step $6$: For any $b<d$, we define

$\mathcal{S}([b,d]):=\sum_{\begin{subarray}{c}1\leq i\leq N,\\\ 1\leq j\leq
m_{i};\\\ b_{i,j}=b\end{subarray}}S_{i}([b_{i,j},d])\;\;\subseteq C_{0}^{K}.$

Notice that

$\dim\left(S_{i}([b_{i,k_{i}},d])+\sum_{j=k_{i}+1}^{m_{i}}S_{i}([b_{i,j},d])\right)=m_{i}-1.$

This follows from the fact the family
$\\{S_{i}([b_{i,k_{i}},d])\\}\cup\\{S_{i}([b_{i,j},d])\\}_{j=k_{i}+1}^{m_{i}}$
is transversal by construction. Moreover, observe that $S_{i_{1}}(I)$ and
$S_{i_{2}}(J)$ are orthogonal to each other whenever $i_{1}\neq i_{2}$ for
every interval $I$ and $J$. Thus, we conclude that $\dim(\mathcal{S}([b,d]))$
is the number of connected components that are born at $b$ and died at $d$.

Recall that we have
$\\{v_{i,1},\ldots,v_{i,l_{i}}\\}=B_{i}\setminus(\cup_{j=1}^{m_{i}}B_{i,j})$.
Let $B^{\mathrm{o}}_{i}:=(\cup_{j=1}^{m_{i}}B_{i,j})$. We define

$\displaystyle S_{i}([d,d])$
$\displaystyle:=\sum_{j=1}^{l_{i}}\operatorname{span}\\{v_{i,j}-c(B^{\mathrm{o}}_{i})\\},$
$\displaystyle\mathcal{S}([d,d])$ $\displaystyle:=\sum_{i=1}^{N}S_{i}([d,d]).$

######  Example D.1.

We illustrate the constructions of $S([b_{i,j},d])$ through an explicit
treegram. Consider the part of a treegram depicted in Fig. 13. In this case,
we have

$\displaystyle B_{i,1}$ $\displaystyle=\\{x,y,z,u\\},$ $\displaystyle B_{i,2}$
$\displaystyle=\\{v,w\\},$ $\displaystyle B_{i,3}$ $\displaystyle=\\{g,h\\},$
$\displaystyle B_{i,4}$ $\displaystyle=\\{k\\},$ $\displaystyle B_{i,5}$
$\displaystyle=\\{l,m,n\\},$ $\displaystyle B_{i,6}$
$\displaystyle=\\{p,q,r\\},$ $\displaystyle B^{\mathrm{o}}_{i}$
$\displaystyle=(\cup_{j=1}^{6}B_{i,j}),$ $\displaystyle B_{i}$
$\displaystyle=B^{\mathrm{o}}_{i}\cup\\{a_{1},a_{2}\\}.$

In this example, it holds that
$b_{1}:=b_{i,1}=b_{i,2}=b_{i,3}=b_{i,4}<b_{i,5}=b_{i,6}=:b_{2}$. Following the
definitions of $R_{i,j}$ and ${R^{\prime}}_{i,j}$ we compute

$\displaystyle R_{i,1}$ $\displaystyle=\\{y\\},$ $\displaystyle R_{i,2}$
$\displaystyle=\\{v,w\\},$ $\displaystyle R_{i,3}$ $\displaystyle=\\{g\\},$
$\displaystyle R_{i,4}$ $\displaystyle=\\{k\\},$ $\displaystyle R_{i,5}$
$\displaystyle=\\{l,n\\},$ $\displaystyle R_{i,6}$
$\displaystyle=\\{p,q,r\\},$ $\displaystyle{R^{\prime}}_{i,5}$
$\displaystyle={R^{\prime}}_{i,6}=\\{x,y,z,v,w,g,h,k\\}.$

Then, we see that
${c^{\prime}}_{i,5}={c^{\prime}}_{i,6}=\frac{1}{8}(x+y+z+v+w+g+h+k)$ and
$c(B^{\mathrm{o}}_{i})=\frac{1}{15}(x+y+z+u+v+w+g+h+k+l+m+n+p+q+r)$. Hence, we
have that

$\displaystyle S_{i}([b_{1},d])$
$\displaystyle=\operatorname{span}\\{\frac{1}{2}(v+w)-y\\}+\operatorname{span}\\{g-y\\}+\operatorname{span}\\{k-y\\},$
$\displaystyle S_{i}([b_{2},d])$
$\displaystyle=\operatorname{span}\\{\frac{1}{2}(l+n)-{c^{\prime}}_{i,5}\\}+\operatorname{span}\\{\frac{1}{3}(p+q+r)-{c^{\prime}}_{i,6}\\},$
$\displaystyle S_{i}([d,d])$
$\displaystyle=\operatorname{span}\\{a_{1}-c(B^{\mathrm{o}}_{i})\\}+\operatorname{span}\\{a_{2}-c(B^{\mathrm{o}}_{i})\\}$

Figure 13: A treegram for illustrating the construction of $\mathcal{S}$; see
Example D.1

######  Proposition D.2.

Let $\mathsf{F}$ be a filtration and let $T_{\mathsf{F}}$ be its treegram. Let
$\mathcal{S}$ be constructed as explained above. Then, for any $b\leq d$,
$\mathcal{S}([b,d])=\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}([b,d])$.

###### Proof.

Assume $b<d$ and let $i,j$ be such that $b_{i,j}=b$. Observe that it is enough
to check
$S_{i}([b_{i,j},d])\subseteq\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d])$.
The equality would then follow from the fact that the sum in the definition of
$\mathcal{S}([b,d])$ is a direct sum by the construction of $S_{i}$s and the
fact that the dimensions of $\mathcal{S}([b,d])$ and
$\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}([b,d])$ have to be the same
as they are both equal to the number of connected components that are born at
time $b$ and die at time $d$, as explained during the construction of
$\mathcal{S}$.

Recall that by Proposition 4.23 we have

$\mathsf{OI}_{\times}\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d])=\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d])\ominus(\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])+\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d-1])).$

So, we need to check that $S_{i}([b_{i,j},d])$ is orthogonal to both
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])$ and
$\mathsf{ZB}_{0}([b_{i,j},d-1])$. There are two cases to be checked, namely
$1\leq j\leq k_{i}$ and $k_{i}+1\leq j\leq m_{i}$. As the processes, in either
case, are similar to each other, we provide details for the case $1\leq j\leq
k_{i}$.

Assume that $1\leq j\leq k_{i}$. In this case, we have that

$S_{i}([b_{i,j},d])=S_{i}([b_{i,k_{i}},d])=\sum_{l=2}^{k}\operatorname{span}\\{c_{i,l}-c_{i,1}\\}.$

Thus, it is enough to check that $c_{i,l}-c_{i,1}$ is orthogonal to both
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])$ and
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d-1])$ for $l=2,\ldots,k_{i}$. The
support of the chain $c_{i,l}-c_{i,1}$ is a subset of $B_{i}$. On the other
hand, support of any chain in $\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])$ is
in the complement of $B_{i}$. Thus, $c_{i,l}-c_{i,1}$ is orthogonal to
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])$. The subspace
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d-1])$ is generated by elements of the
form $x-y$ where $x$ and $y$ belong to the same block of
$T_{\mathsf{F}}(d-1)$. Thus, either support of $c_{i,l}-c_{i,1}$ and the set
$\\{x,y\\}$ are disjoint or $x$ and $y$ are both in the support of
$c_{i,l}-c_{i,1}$ with the same coefficients. In both scenarios, we get that
$c_{i,l}-c_{i,1}$ is orthogonal to
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d-1])$. Thus, $c_{i,l}-c_{i,1}$ is
orthogonal to $\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j}-1,d])$ and
$\mathsf{ZB}_{0}^{\mathsf{F}}([b_{i,j},d-1])$.

When $b=d$, the proof is similar. ∎

##  Appendix E Harmonic persistent homology and Grassmannian Persistence
Diagrams

In this section, we revisit some definitions and results from the work of Basu
and Cox in [2]. We establish a relation between their construction: _Harmonic
barcodes_ , and our construction: Grassmannian persistence diagrams.

Let $\mathsf{F}:=\\{K_{i}\\}_{i=1}^{n}$ be a filtration of a finite simplicial
complex $K$ ($K_{n}=K$) over the poset $[n]:=\\{1<\cdots<n\\}$. In [2], for
every degree $q\geq 0$, the authors assign a vector space (subspace of the
$q$-th chain space of $K$) to every interval $[i,j]$ for $i<j$. In Section 5,
we also assign a vector space (also a subspace of the $q$-th chain space of
$K$) to every interval $[i,j]$ for $i<j$ (i.e. Grassmannian persistence
diagram). In both scenarios, for every interval $[i,j]$, the multiplicity of
the interval $[i,j]$ in the degree-$q$ persistence diagram of the filtration
$F$ is the same as the dimension of the vector space assigned to that
interval. Therefore, the two subspaces assigned to a point in the persistence
diagram are isomorphic, as their dimensions are the same. However, they are
not the same subspace of the chain space. In this section, we construct an
explicit isomorphism between these two subspaces through a certain projection.

We now recall the definition of _persistent homology group_ from [9] and some
related definition from [2].

######  Definition E.1 (Persistent homology group [2, Definitions 8, 10, 11]).

Let $\mathsf{F}=\\{K_{i}\\}_{i=1}^{n}$ be a filtration. Let
$\iota_{q}^{i,j}:H_{q}(K_{i})\to H_{q}(K_{j})$ denote the homomorpshims
induced by the inclusion $K_{i}\hookrightarrow K_{j}$. The  _persistent
homology group_ , $H_{q}^{i,j}(\mathsf{F})$, of $\mathsf{F}$ is defined by

$H_{q}^{i,j}(\mathsf{F}):=\operatorname{im}(\iota_{q}^{i,j}).$

For $i<j$, also define

$\displaystyle M_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\iota_{q}^{i,j})^{-1}(H_{q}^{i-1,j}(\mathsf{F}))$
$\displaystyle N_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\iota_{q}^{i,j-1})^{-1}(H_{q}^{i-1,j-1}(\mathsf{F}))$
$\displaystyle P_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=M_{q}^{i,j}(\mathsf{F})/N_{q}^{i,j}(\mathsf{F})$

where $K_{0}$ is taken to be equal to $K_{1}$ by convention.

######  Remark E.2.

As noted in [2, Remark 4], the interpretation of these subspaces are as
follows:

  * •

$M_{q}^{i,j}(\mathsf{F})$ is a subspace of $H_{q}(K_{i})$ consisting of
homology classes in $H_{q}(K_{i})$ which are

“born before $i$, or born at $i$ and dies at $j$ or earlier.”

  * •

$N_{q}^{i,j}(\mathsf{F})$ is a subspace of $H_{q}(K_{i})$ consisting of
homology classes in $H_{q}(K_{i})$ which are

“born before $i$, or born at $i$ and dies strictly earlier than $j$.”

  * •

$P_{q}^{i,j}(\mathsf{F})$ is the space of $q$-dimensional cycles that are

“born at $i$ and dies at $j$.”

######  Definition E.3 (Harmonic homology space [2, Definition 5]).

The  _harmonic homology space_ of $K$ is the subspace
$\mathcal{H}_{q}(K)\subseteq C_{q}^{K}$ defined by

$\mathcal{H}_{q}(K):=\mathsf{Z}_{q}(K)\cap\mathsf{B}_{q}(K)^{\perp},$

where $C_{q}^{K}$ is endowed with the standard inner product as described in
Section 2.

######  Proposition E.4 ([2, Proposition 2.5]).

The map $\mathfrak{f}_{q}(K):H_{q}(K)\to\mathcal{H}_{q}(K)$ defined by

$z+\mathsf{B}_{q}(K)\mapsto\operatorname{proj}_{(\mathsf{B}_{q}(K))^{\perp}}(z)$

is an isomorphism of vector spaces.

######  Proposition E.5 ([2, Proposition 1]).

Let $\mathsf{F}=\\{K_{i}\\}_{i=1}^{n}$ be a filtration. For $i\leq j$, the
restriction of $\operatorname{proj}_{(\mathsf{B}_{q}(K_{j}))^{\perp}}$ to
$\mathcal{H}_{q}(K_{i})$ gives a linear map

$\gamma_{q}^{i,j}:=\left.\left(\operatorname{proj}_{(\mathsf{B}_{q}(K_{j}))^{\perp}}\right)\right|_{\mathcal{H}_{q}(K_{i})}:\mathcal{H}_{q}(K_{i})\to\mathcal{H}_{q}(K_{j})$

which makes the following diagram commute

${H_{q}(K_{i})}$${H_{q}(K_{j})}$${\mathcal{H}_{q}(K_{i})}$${\mathcal{H}_{q}(K_{j})}$$\scriptstyle{\iota_{q}^{i,j}}$$\scriptstyle{\mathfrak{f}_{q}(K_{i})}$$\scriptstyle{\mathfrak{f}_{q}(K_{j})}$$\scriptstyle{\gamma_{q}^{i,j}}$

######  Definition E.6 (Harmonic persistent homology group [2, Definitions 13
and 14]).

Let $\mathsf{F}=\\{K_{i}\\}_{i=1}^{n}$ be a filtration. Let
$\gamma_{q}^{i,j}:\mathcal{H}_{q}(K_{i})\to\mathcal{H}_{q}(K_{j})$ denote the
maps defined in Proposition E.5. The  _harmonic persistent homology group_ ,
$\mathcal{H}_{q}^{i,j}(\mathsf{F})$, of $\mathsf{F}$ is defined by

$\mathcal{H}_{q}^{i,j}(\mathsf{F}):=\operatorname{im}(\gamma_{q}^{i,j}).$

For $i<j$, also define

$\displaystyle\mathcal{M}_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\gamma_{q}^{i,j})^{-1}(\mathcal{H}_{q}^{i-1,j}(\mathsf{F}))$
$\displaystyle\mathcal{N}_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\gamma_{q}^{i,j-1})^{-1}(\mathcal{H}_{q}^{i-1,j-1}(\mathsf{F}))$
$\displaystyle\mathcal{P}_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=\mathcal{M}_{q}^{i,j}(\mathsf{F})\cap(\mathcal{N}_{q}^{i,j}(\mathsf{F}))^{\perp}$

where $K_{0}$ is taken to be equal to $K_{1}$ by convention. The map

$\displaystyle\mathsf{HB}_{q}^{\mathsf{F}}:\overline{[n]}^{\supseteq}$
$\displaystyle\to\mathsf{Gr}(C_{q}^{K})$ $\displaystyle[i,j]$
$\displaystyle\mapsto\mathcal{P}_{q}^{i,j}(\mathsf{F})$

is called the degree-$q$ _Harmonic barcode_ of $\mathsf{F}$.

We now establish the connection between the degree-$q$ Grassmannian
persistence diagram and the degree-$q$ Harmonic barcode of a filtration
$\mathsf{F}$.

######  Theorem E.7.

Let $\mathsf{F}:=\\{K_{i}\\}_{i=1}^{n}$ be a filtration. For $i<j$,

$\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}\right)^{\perp}}:\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}(i,j)\to\mathcal{P}_{q}^{i,j}(\mathsf{F})$

is an isomorphism.

Note that the dimensions of the vector spaces
$\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}(i,j)$ and
$\mathcal{P}_{q}^{i,j}(\mathsf{F})$ are the same as both are equal to the
multiplicity of the interval $[i,j]$ in the degree-$q$ persistence diagram of
$\mathsf{F}$. So, it is already known that
$\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}(i,j)$ and
$\mathcal{P}_{q}^{i,j}(\mathsf{F})$ are isomorphic. Our Theorem E.7 shows that
this isomorphism can be written explicitly as a projection.

######  Definition E.8.

For a simplicial complex $K$ and any degree $q\geq 0$, let

$\phi_{q}^{K}:\mathsf{Z}_{q}(K)\to\mathsf{Z}_{q}(K)/\mathsf{B}_{q}(K)=H_{q}(K)$

denote the canonical quotient map. For a filtration
$\mathsf{F};=\\{K_{i}\\}_{i=1}^{n}$, we define, for $i<j$,

$\displaystyle\tilde{M}_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\phi_{q}^{K_{i}})^{-1}(M_{q}^{i,j}(\mathsf{F}))$
$\displaystyle\tilde{N}_{q}^{i,j}(\mathsf{F})$
$\displaystyle:=(\phi_{q}^{K_{i}})^{-1}(N_{q}^{i,j}(\mathsf{F})).$

######  Lemma E.9.

$\displaystyle\tilde{M}_{q}^{i,j}(\mathsf{F})$
$\displaystyle=\\{z\in\mathsf{Z}_{q}(K_{i})\mid\exists
z_{i-1}\in\mathsf{Z}_{q}(K_{i-1})\text{ such that
}z-z_{i-1}\in\mathsf{B}_{q}(K_{j})\\}$
$\displaystyle\tilde{N}_{q}^{i,j}(\mathsf{F})$
$\displaystyle=\\{z\in\mathsf{Z}_{q}(K_{i})\mid\exists
z_{i-1}\in\mathsf{Z}_{q}(K_{i-1})\text{ such that
}z-z_{i-1}\in\mathsf{B}_{q}(K_{j-1})\\}$

###### Proof.

As
$\tilde{M}_{q}^{i,j}(\mathsf{F})=(\phi_{q}^{K_{i}})^{-1}(M_{q}^{i,j}(\mathsf{F}))$,
we start by computing $M_{q}^{i,j}(\mathsf{F})$ as follows

$\displaystyle M_{q}^{i,j}(\mathsf{F})$
$\displaystyle=(\iota_{q}^{i,j})^{-1}(H_{q}^{i-1,j}(\mathsf{F}))$
$\displaystyle=(\iota_{q}^{i,j})^{-1}\left(\frac{\mathsf{Z}_{q}(K_{i-1})+\mathsf{B}_{q}(K_{j})}{\mathsf{B}_{q}(K_{j})}\right)$
$\displaystyle=\left\\{z+\mathsf{B}_{q}(K_{i})\in H_{q}(K_{i})\mid
z+\mathsf{B}_{q}(K_{j})\in\frac{\mathsf{Z}_{q}(K_{i-1})+\mathsf{B}_{q}(K_{j})}{\mathsf{B}_{q}(K_{j})}\right\\}$
$\displaystyle=\\{z+\mathsf{B}_{q}(K_{i})\in H_{q}(K_{i})\mid\exists
z_{i-1}\in\mathsf{Z}_{q}(K_{i-1})\text{ such that
}z-z_{i-1}\in\mathsf{B}_{q}(K_{j})\\}$

Therefore, we have that

$\displaystyle\tilde{M}_{q}^{i,j}(\mathsf{F})$
$\displaystyle=(\phi_{q}^{K_{i}})^{-1}(M_{q}^{i,j}(\mathsf{F}))$
$\displaystyle=\\{z\in\mathsf{Z}_{q}(K_{i})\mid\exists
z_{i-1}\in\mathsf{Z}_{q}(K_{i-1})\text{ such that
}z-z_{i-1}\in\mathsf{B}_{q}(K_{j})\\}$

The result for $\tilde{N}_{q}^{i,j}(\mathsf{F})$ follows from the fact that
$N_{q}^{i,j}(\mathsf{F})=M_{q}^{i,j-1}(\mathsf{F})$. ∎

######  Lemma E.10.

The map

$\displaystyle\varphi:\frac{\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])}{\mathsf{ZB}_{q}(i-1,j)+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1])}$
$\displaystyle\to\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})}{\tilde{N}_{q}^{i,j}(\mathsf{F})}$
$\displaystyle
z+\left(\mathsf{ZB}_{q}^{\mathsf{F}}([i-1,j])+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1])\right)$
$\displaystyle\mapsto z+\tilde{N}_{q}^{i,j}(\mathsf{F})$

is an isomorphism.

###### Proof.

Observe that
$\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])\subseteq\tilde{M}_{q}^{i,j}(\mathsf{F})$.
This suggests defining the map

$\displaystyle\psi:\mathsf{ZB}_{q}(i,j)$
$\displaystyle\to\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})}{\tilde{N}_{q}^{i,j}(\mathsf{F})}$
$\displaystyle z$ $\displaystyle\mapsto z+\tilde{N}_{q}^{i,j}(\mathsf{F}).$

Then, using the description of $\tilde{M}_{q}^{i,j}(\mathsf{F})$ and
$\tilde{N}_{q}^{i,j}(\mathsf{F})$ given in Lemma E.9, it is straightforward to
check that $\psi$ is surjective and that

$\ker(\psi)=\mathsf{ZB}_{q}^{\mathsf{F}}([i-1,j])+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1]),$

which concludes the proof. ∎

###### Proof of Theorem E.7.

Notice that by Proposition 4.23 we have

$\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])=\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])\ominus\left(\mathsf{ZB}_{q}^{\mathsf{F}}([i-1,j])+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1])\right).$

Furthermore, the map

$\displaystyle[\cdot]_{\mathsf{ZB}_{q}}:\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}$
$\displaystyle\to\frac{\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])}{\left(\mathsf{ZB}_{q}^{\mathsf{F}}([i-1,j])+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1])\right)}$
$\displaystyle z$
$\displaystyle\mapsto[z]_{\mathsf{ZB}_{q}}:=z+\left(\mathsf{ZB}_{q}^{\mathsf{F}}(i-1,j)+\mathsf{ZB}_{q}^{\mathsf{F}}(i,j-1)\right)$

is an isomorphism. Let $\varphi$ be the isomorphism from the Lemma E.10. Let

$\theta:\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})}{\tilde{N}_{q}^{i,j}(\mathsf{F})}\to\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})/\mathsf{B}_{q}(K_{i})}{\tilde{N}_{q}^{i,j}(\mathsf{F})/\mathsf{B}_{q}(K_{i})}$

be the canonical isomorphism. Let

$\mu_{q}(K_{i}):\frac{M_{q}^{i,j}(\mathsf{F})}{N_{q}^{i,j}(\mathsf{F})}\to\frac{\mathcal{M}_{q}^{i,j}(\mathsf{F})}{\mathcal{N}_{q}^{i,j}(\mathsf{F})}$

be defined by

$\mu_{q}(K_{i})((z+\mathsf{B}_{q}(K_{i}))+N_{q}^{i,j}(\mathsf{F})):=\operatorname{proj}_{(\mathsf{B}_{q}(K_{i}))^{\perp}}(z)+\mathcal{N}_{q}^{i,j}(\mathsf{F})$

for
$(z+\mathsf{B}_{q}(K_{i}))+N_{q}^{i,j}(\mathsf{F})\in\frac{M_{q}^{i,j}(\mathsf{F})}{N_{q}^{i,j}(\mathsf{F})}$.
By Proposition E.5, $\mu_{q}(K_{i})$ is a well-defined isomorphism. Finally,
we have the isomorphism

$\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}(\mathsf{F})\right)^{\perp}}:\frac{\mathcal{M}_{q}^{i,j}(\mathsf{F})}{\mathcal{N}_{q}^{i,j}(\mathsf{F})}\to\mathcal{M}_{q}^{i,j}(\mathsf{F})\cap(\mathcal{N}_{q}^{i,j}(\mathsf{F}))^{\perp}$

Now, composing all these isomorphisms as in the diagram below

${{\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])}}$${{\frac{\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])}{\left(\mathsf{ZB}_{q}^{\mathsf{F}}([i-1,j])+\mathsf{ZB}_{q}^{\mathsf{F}}([i,j-1])\right)}}}$${{\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})}{\tilde{N}_{q}^{i,j}(\mathsf{F})}}}$${{\frac{\mathcal{M}_{q}^{i,j}(\mathsf{F})}{\mathcal{N}_{q}^{i,j}(\mathsf{F})}}}$${{\frac{M_{q}^{i,j}(\mathsf{F})}{N_{q}^{i,j}(\mathsf{F})}}}$${{\frac{\tilde{M}_{q}^{i,j}(\mathsf{F})/\mathsf{B}_{q}(K_{i})}{\tilde{N}_{q}^{i,j}(\mathsf{F})/\mathsf{B}_{q}(K_{i})}}}$${{\mathcal{M}_{q}^{i,j}(\mathsf{F})\cap(\mathcal{N}_{q}^{i,j}(\mathsf{F}))^{\perp}=\mathcal{P}_{q}^{i,j}(\mathsf{F})}}$$\scriptstyle{[\cdot]_{\mathsf{ZB}_{q}}}$$\scriptstyle{\varphi}$$\scriptstyle{\theta}$$\scriptstyle{\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}(\mathsf{F})\right)^{\perp}}}$$\scriptstyle{\mu_{q}(K_{i})}$$\scriptstyle{=}$

we obtain the isomorphism

$\displaystyle\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])$
$\displaystyle\to\mathcal{P}_{q}^{i,j}(\mathsf{F})$ $\displaystyle z$
$\displaystyle\mapsto\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}(\mathsf{F})\right)^{\perp}}\circ\operatorname{proj}_{\left(\mathsf{B}_{q}(K_{i})\right)^{\perp}}(z).$

Observe that, as
$z\in\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}([i,j])$, we have that
$z\in(\mathsf{B}_{q}(K_{i}))^{\perp}$. Therefore,
$\operatorname{proj}_{\left(\mathsf{B}_{q}(K_{i})\right)^{\perp}}(z)=z$. Thus,
the isomorphism above is given by

$\displaystyle\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}\right)^{\perp}}:\mathsf{OI}_{\times}\mathsf{ZB}_{q}^{\mathsf{F}}(i,j)$
$\displaystyle\to\mathcal{P}_{q}^{i,j}(\mathsf{F})$ $\displaystyle z$
$\displaystyle\mapsto\operatorname{proj}_{\left(\mathcal{N}_{q}^{i,j}(\mathsf{F})\right)^{\perp}}(z)$

as stated in Theorem E.7. ∎