$e\in E(M)$ is a loop then $\partial_{e}$ is clearly in the annihilator. This
claim follows directly from the fact that $\operatorname{A}(M)$ is a quotient
ring of the graded Möbius algebra. Since the graded Möbius algebra only
depends on the lattice of flats which only depends on the simplification of
the matroid, it follows that $\operatorname{A}(M)$ should only depend on the
simplification of the matroid. However, we want to prove a more delicate
isomorphism which tells us the image of linear forms under this isomorphism.
This is the content of Theorem 4.23. Before proving this result, we first
describe some common elements in the ideal $\operatorname{Ann}_{M}$.

######  Proposition 4.22.

Let $M=(E,\mathcal{I})$ be a matroid. For any subsets $S,T\subseteq E$, we
write $S\sim T$ if and only if $\overline{S}=\overline{T}$. Let
$\Lambda_{M}^{(1)},\Lambda_{M}^{(2)}$, $\Lambda_{M}^{(3)}$ be three subsets of
$\operatorname{S}_{M}$ given by

$\displaystyle\Lambda_{M}^{(1)}$ $\displaystyle:=\left\\{X_{e}^{2}:e\in
E(M)\right\\},$ $\displaystyle\Lambda_{M}^{(2)}$
$\displaystyle:=\left\\{X^{S}:S\notin\mathcal{I}(M)\right\\},$
$\displaystyle\Lambda_{M}^{(3)}$
$\displaystyle:=\left\\{X^{S}-X^{T}:S,T\in\mathcal{I}(M)\text{ such that
}S\sim T\right\\}.$

Let $\Lambda_{M}=\Lambda_{M}^{(1)}\cup\Lambda_{M}^{(2)}\cup\Lambda_{M}^{(3)}$.
Then $\Lambda_{M}\subseteq\operatorname{Ann}_{M}$.

###### Proof.

See Proposition 3.1 in [37] ∎

In general, it is not true that the elements in $\Lambda_{M}$ generate the
annihilator $\operatorname{Ann}_{M}$. In fact, if we let $I(\Lambda_{M})$ be
the ideal generated by $\Lambda_{M}$, then
$\operatorname{S}_{M}/I(\Lambda_{M})$ is exactly the graded Möbius algebra
$\operatorname{H}(M)$. The graded Möbius algebra is in general not even a
Poincaré duality algebra, hence it cannot be $\operatorname{A}(M)$. We can
find an explicit counterexample in [37]. Consider the matrix given by

$A=\begin{bmatrix}1&0&0&1&0\\\ 0&1&0&1&1\\\ 0&0&1&0&1\end{bmatrix}.$

The basis of the linear matroid generated by $A$ are
$\\{123,125,134,135,145\\}$. From Example 3.5 in [37], we have that

$\operatorname{Ann}_{M}=I(\Lambda_{M}\cup(X_{1}X_{3}+X_{4}X_{5}-X_{1}X_{5}-X_{3}X_{4})).$

Let $M=(E,\mathcal{I})$ be a matroid and $\widetilde{M}$ be its
simplification. We can define the maps
$\phi:\operatorname{S}_{M}\to\operatorname{S}_{\widetilde{M}}$ and
$\psi:\operatorname{S}_{\widetilde{M}}\to\operatorname{S}_{M}$ by

$\displaystyle\phi(\partial_{x_{e}}):=\partial_{\overline{x_{e}}},\quad\text{and}\quad\psi(\partial_{\overline{x}}):=\frac{1}{|\overline{x}|}\sum_{e\in\overline{x}}\partial_{x_{e}}.$

and then extending to the whole polynomial ring using the universal property
of polynomial rings.

######  Theorem 4.23.

The maps $\phi:S_{M}\to S_{\widetilde{M}}$ and $\psi:S_{\widetilde{M}}\to
S_{M}$ induce isomorphisms between $A(M)$ and $A(\widetilde{M})$.

###### Proof.

We first prove that $\phi$ and $\psi$ induce homomorphisms between the rings
$\operatorname{A}(M)$ and $\operatorname{A}(\widetilde{M})$. To show that
$\phi$ induces a homomorphism, consider the diagram in Equation 4.2.

${{S_{M}}}$${{S_{\widetilde{M}}}}$${{A(\widetilde{M})}}$${{A(M)}}$$\scriptstyle{\phi}$$\scriptstyle{\pi_{\widetilde{M}}}$$\scriptstyle{\pi_{M}}$$\scriptstyle{\exists!\Phi}$
(4.2)

Let $\xi\in S_{M}$ be an element satisfying $\xi(f_{M})=0$. We will prove that
$\phi(\xi)(f_{\widetilde{M}})=0$. In other words, we want to prove that
$\operatorname{Ann}_{M}\subseteq\ker\pi_{\widetilde{M}}\circ\phi$. From
Proposition 4.8, it suffices to consider the case where $\xi$ is homogeneous.
Let $e_{1},\ldots,e_{s}$ be representatives of all parallel classes. Then, we
have that $M=\overline{e_{1}}\sqcup\ldots\sqcup\overline{e_{s}}\cup E_{0}$
where $E_{0}$ denotes the loops in $M$. In terms of the basis generating
polynomials, we have

$\displaystyle
f_{\widetilde{M}}(x_{\overline{e_{1}}},\ldots,x_{\overline{e_{s}}})$
$\displaystyle=\sum_{\begin{subarray}{c}1\leq i_{1}<\ldots<i_{r}\leq s\\\
\\{\overline{e_{i_{1}}},\ldots,\overline{e_{i_{r}}}\\}\in\mathcal{B}(\widetilde{M})\end{subarray}}x_{\overline{e_{i_{1}}}}\ldots
x_{\overline{e_{i_{r}}}}$ $\displaystyle f_{M}(x_{1},\ldots,x_{n})$
$\displaystyle=f_{\widetilde{M}}\left(y_{1},\ldots,y_{s}\right)$

where for $1\leq i\leq s$, we define
$y_{i}:=\sum_{e\in\overline{e_{i}}}x_{e}$. Since $\xi$ is homogeneous of
degree $k$, we can write it in the form
$\xi=\sum_{\begin{subarray}{c}\alpha\subseteq[n]\\\
|\alpha|=k\end{subarray}}c_{\alpha}\partial^{\alpha}$. Then, we have

$\displaystyle\xi(f_{M})$
$\displaystyle=\sum_{\beta\in\mathcal{B}}\xi(x^{\beta})=\sum_{\beta\in\mathcal{B}(M)}\sum_{\begin{subarray}{c}\alpha\subseteq[n]\\\
|\alpha|=k\end{subarray}}c_{\alpha}\partial^{\alpha}x^{\beta}=\sum_{\gamma\in\mathcal{I}_{r-k}(M)}\left(\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}\\\
\alpha\cup\gamma\in\mathcal{I}_{r}(M)\end{subarray}}c_{\alpha}\right)x^{\gamma}.$
(4.3)

Since $\xi(f_{M})=0$, we know that all of the coefficients on the right hand
side of Equation 4.3 are equal to $0$. Thus, we have that

$\sum_{\begin{subarray}{c}\alpha\in I_{k}\\\
\alpha\cup\gamma\in\mathcal{I}_{r}(M)\end{subarray}}c_{\alpha}=0,\text{ for
all $\gamma\in\mathcal{I}_{r-k}(M)$}.$

On the other hand, we have

$\phi(\xi)=\sum_{\begin{subarray}{c}\alpha\subseteq[n]\\\
|\alpha|=k\end{subarray}}c_{\alpha}\prod_{e\in\alpha}\partial_{\overline{e}}=\sum_{\beta\in\mathcal{I}_{k}(\widetilde{M})}\left(\sum_{\alpha\in\text{fiber}(\beta)}c_{\alpha}\right)\partial^{\beta}.$

We can let this differential act on $f_{\widetilde{M}}$ to get the expression

$\displaystyle\phi(\xi)(f_{\widetilde{M}})$
$\displaystyle=\sum_{\gamma\in\mathcal{I}_{r-k}(\widetilde{M})}\left(\sum_{\begin{subarray}{c}\beta\in\mathcal{I}_{k}(\widetilde{M})\\\
\beta\cup\gamma\in\mathcal{B}(\widetilde{M})\end{subarray}}\sum_{\alpha\in\text{fiber}(\beta)}c_{\alpha}\right)x^{\gamma}=\sum_{\gamma\in\mathcal{I}_{r-k}(\widetilde{M})}\left(\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}(M)\\\
\alpha\cup\gamma_{0}\in\mathcal{B}(M)\end{subarray}}c_{\alpha}\right)x^{\gamma}=0.$
(4.4)

In Equation 4.4, the independent set $\gamma_{0}\in\text{fiber}(\gamma)$ is an
arbitrary element in the fiber of $\gamma$. This proves that
$\operatorname{Ann}_{M}\subseteq\ker\pi_{\widetilde{M}}\circ\phi$. Thus, there
is a unique ring homomorphism $\Phi:A(M)\to A(\widetilde{M})$ which makes
Equation 4.2 commute.

To prove that $\psi$ induces a ring homomorphism, consider the diagram in
Equation 4.5.

${{S_{\widetilde{M}}}}$${{S_{M}}}$${{A(M)}}$${{A(\widetilde{M})}}$$\scriptstyle{\pi_{\widetilde{M}}}$$\scriptstyle{\psi}$$\scriptstyle{\pi_{M}}$$\scriptstyle{\exists!\Psi}$
(4.5)

Consider a differential $\xi\in S_{\widetilde{M}}$ satisfying
$\xi(f_{\widetilde{M}})=0$. We can write this as
$\xi=\sum_{\alpha\in\mathcal{I}_{k}(\widetilde{M})}c_{\alpha}\partial^{\alpha}$
for some real constants $c_{\alpha}$. Then, its image under $\psi$ is equal to

$\psi(\xi)=\sum_{\alpha\in\mathcal{I}_{k}(\widetilde{M})}\frac{c_{\alpha}}{\prod_{e\in\alpha}|e|}\sum_{\beta\in\text{fiber}(\alpha)}\partial^{\beta}.$

Fix a $\alpha\in\mathcal{I}_{k}(\widetilde{M})$ and a
$\beta\in\text{fiber}(\alpha)$. Since $\partial y_{i}/\partial
x_{e}=\mathds{1}_{e\in\overline{e_{i}}}$, we have

$\displaystyle\partial^{\beta}f_{M}(x_{1},\ldots,x_{n})$
$\displaystyle=\partial^{\beta}f_{\widetilde{M}}(y_{1},\ldots,y_{s})=\partial^{\alpha}f_{\widetilde{M}}(x_{1},\ldots,x_{s})|_{x_{1}=y_{1},\ldots,x_{s}=y_{s}}=0.$

Thus, we have that $\psi(\xi)(f_{M})=0$ and
$\operatorname{Ann}_{\widetilde{M}}\subseteq\ker\pi_{M}\circ\psi$. This proves
that there is a unique ring homomrphism $\Psi:A(\widetilde{M})\to A(M)$ which
causes the diagram in Equation 4.5 to commute. Since the maps $\Psi$ and
$\Phi$ are inverses of each other, they are both isomorphisms. This suffices
for the proof. ∎

From Theorem 4.23, we get Corollary 4.24 and Corollary 4.25 immediately.

######  Corollary 4.24.

Let $M=(E,\mathcal{I})$ be a matroid. For any $a\in\mathbb{R}^{E}$, we can
define the linear form $l_{a}:=\sum_{e\in E}a_{e}\cdot\partial_{x_{e}}\in
A^{1}(M)$. Let $\widetilde{l_{a}}:=\Phi(l_{a})=\sum_{e\in
E}a_{e}\cdot\partial_{x_{\overline{e}}}\in A^{1}(\widetilde{M})$. Then, the
following diagram commutes:

${\ldots}$${{A^{i-1}(M)}}$${{A^{i}(M)}}$${{A^{i+1}(M)}}$${\ldots}$${\ldots}$${{A^{i-1}(\widetilde{M})}}$${{A^{i}(\widetilde{M})}}$${{A^{i+1}(\widetilde{M})}}$${\ldots}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\times\widetilde{l_{a}}}$$\scriptstyle{\Phi}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\times\widetilde{l_{a}}}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\times\widetilde{l_{a}}}$$\scriptstyle{\Phi}$$\scriptstyle{\Phi}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\times\widetilde{l_{a}}}$

######  Corollary 4.25.

Let $M=(E,\mathcal{I})$ be a matroid. Then $A(M)$ satisfies $\mathsf{HRR}_{k}$
with respect to $l$ if and only if $A(\widetilde{M})$ satisfies
$\mathsf{HRR}_{k}$ with respect to $\Phi(l)$.

From Theorem 4.23, Corollary 4.24, and Corollary 4.25, when we study Hodge-
theoretic properties of $\operatorname{A}(M)$ it is enough to assume that
$\widetilde{M}$ is simple. We may not always want to do this, but the
translation from a matroid to its simplification is useful nonetheless. Not
only do we know that the two rings are isomorphic, but we know how
multiplication by $1$-forms translate from one ring to the other. This means
that we can relate the Hodge theoretic properties of the two rings to each
other. We also understand some properties of $\operatorname{A}(M)$ better when
we know that $M$ is simple. For example, the vector space structure of
$\operatorname{A}^{1}(M)$ is well-understood for simple $M$. Satoshi Murai,
Takahiro Nagaoka, and Akiko Yazawa prove in [39] that if $M$ is a simple
matroid on $[n]$, then $\dim A^{1}(M)=n$. In other words, the spanning set
$\partial_{1},\ldots,\partial_{n}$ is a basis of $\operatorname{A}^{1}(M)$. We
write this result in Lemma 4.26.

######  Lemma 4.26.

If $M=([n],\mathcal{I})$ is simple, then $\partial_{1},\ldots,\partial_{n}$ is
a basis of $A^{1}(M)$.

###### Proof.

See Theorem 2.5 in [39] ∎

From Corollary 4.20, this implies that when $M$ is simple, the Hessian of the
basis generating polynomial $f_{M}$ has signature $(+,-,\ldots,-)$ at every
point in the positive orthant. Thus, the basis generating polynomial for a
simple matroid is strictly log-concave on the positive orthant, which was one
of the main results in [39]. In the same paper, they formulate Conjecture
4.27. One of my goals was to work towards a proof of Conjecture 4.28.

######  Conjecture 4.27.

Let $M=(E,\mathcal{I})$ be a matroid of rank $d$. The ring
$\operatorname{A}(M)$ satisfies $\mathsf{HL}_{k}$ for some
$a\in\mathbb{R}^{E}_{>0}$ for all $k\leq\frac{d}{2}$.

######  Conjecture 4.28.

Let $M=(E,\mathcal{I})$ be a matroid of rank $d$. The ring
$\operatorname{A}(M)$ satisfies the Kähler package for all
$a\in\mathbb{R}^{E}_{>0}$.

Unfortunately, Conjecture 4.28 is false. The counterexample was given to us
through private communication between June Huh and Connor Simpson. We discuss
the counterexample in more detail in Section 4.6. For the rest of the thesis,
we analyze when $\operatorname{A}(M)$ satisfies $\mathsf{HRR}_{1}$ on the
facets of the positive orthant. We will resolve this question completely in
Section 4.5. Before moving on to this question, we first give necessary and
sufficient conditions on $\operatorname{A}(M)$ for $\mathsf{HRR}_{1}$ with
respect to $a\in\mathbb{R}^{n}$ in the case where $f(a)>0$ and $M$ is simple.

######  Corollary 4.29.

Let $M=(E,\mathcal{I})$ be a simple matroid of rank $r\geq 2$. If
$a\in\mathbb{R}_{\geq 0}^{E}$ satisfies $f_{M}(a)>0$, then $A(M)$ satisfies
$\mathsf{HRR}_{1}$ with respect to $l_{a}$ if and only if
$\operatorname{Hess}_{f_{M}}|_{x=a}$ is non-singular.

###### Proof.

This follows immediately from Lemma 4.16, Corollary 4.20, and Lemma 4.26. ∎

######  Lemma 4.30.

When $D$ is invertible and $A$ is a square matrix, we have

$\det\begin{bmatrix}A&B\\\ C&D\end{bmatrix}=\det(A-BD^{-1}C)\det(D).$

###### Proof.

See Section 5 in [55]. ∎

######  Theorem 4.31.

Let $M=(E,\mathcal{I})$ be a simple matroid of rank $r\geq 2$. If
$a\in\mathbb{R}_{\geq 0}^{E}$ satisfies $f_{M}(a)>0$ and $a_{e}=0$ for some
$e\in E$ which is not a co-loop, then $A(M)$ satisfies $\mathsf{HRR}_{1}$ with
respect to $l_{a}$ if and only if

$\left(\nabla
f_{M/e}^{\mathsf{T}}\cdot\operatorname{Hess}^{-1}_{f_{M\backslash
e}}\cdot\nabla f_{M/e}\right)|_{x=a}\neq 0.$

###### Proof.

Without loss of generality, we can assume that $E(M)=[n]$ and $e=n$. In
particular, this means that $a=(a_{1},\ldots,a_{n-1},0)\in\mathbb{R}^{n}$ with
$a_{i}\geq 0$ for all $i$. From Corollary 4.29, $\mathsf{HRR}_{1}$ is
satisfied if and only if the Hessian is non-singular. To compute the Hessian
at $x=a$, note that because $n$ is a co-loop, we can write the basis
generating polynomial as $f_{M}=x_{n}f_{M/n}+f_{M\backslash n}$. Using this
equation, we see that the Hessian of $f_{M}$ at $(a_{1},\ldots,a_{n-1},0)$ is
equal to

$\operatorname{Hess}_{f_{M}}=\begin{bmatrix}\operatorname{Hess}_{f_{M\backslash
n}}&\nabla f_{M/n}\\\ (\nabla f_{M/n})^{\mathsf{T}}&0\end{bmatrix}.$

Since $M$ is simple, we know that $M\backslash n$ is simple. Thus, the matrix
$\operatorname{Hess}_{f_{M\backslash n}}$ is invertible as it has the same
signature as the Hodge-Riemann form. From the Lemma 4.30, we have that

$\det\operatorname{Hess}_{f_{M}}=\left(\nabla
f_{M/n}^{\mathsf{T}}\cdot\operatorname{Hess}^{-1}_{f_{M\backslash
n}}\cdot\nabla f_{M/n}\right).$

This suffices for the proof. ∎

###  4.5 Hodge-Riemann Relations on the Facets of the Positive Orthant

In this section, we will prove necessary and sufficient conditions for
$\operatorname{A}(M)$ to satisfy $\mathsf{HRR}_{1}$ on the facets of the
positive orthant. From Lemma 4.19, we already know that $A(M)$ satisfies
$\mathsf{HRR}_{1}$ on $\mathbb{R}_{>0}^{n}$. For a matroid
$M=(E,\mathcal{I})$, we can decompose the boundary set of
$\mathbb{R}^{E}_{\geq 0}$ as

$\operatorname{bd}\mathbb{R}_{\geq 0}^{E}=\bigcup_{e\in E}H_{e}$

where $H_{e}:=\\{x\in\mathbb{R}_{\geq 0}^{E}:x_{e}=0\\}$. Before stating our
main result in this section, we first prove a few technical lemmas that are
necessary for us to apply the inductive procedure modeled in Lemma 4.18.

######  Lemma 4.32.

Let $M=(E,\mathcal{I})$ be a matroid satisfying $\operatorname{rank}(M)\geq
2$. If $e\in M$ is not a coloop of $M$, then $e$ will not be a co-loop of
$M/i$ for any $i\in E\backslash e$.

###### Proof.

If $e$ or $i$ is a loop, then the statement is vacuously true. Now, suppose
that $e$ and $i$ are both not loops. Suppose for the sake of contradiction
that $e$ is a coloop of $M/i$. This implies that any basis of $M$ containing
$i$ must contain $e$. But, since $e$ is not a coloop of $M$, the matroid
$M\backslash e$ has the same rank of $M$. Moreover, since $i$ is independent
in $M\backslash e$, there is a basis of $M\backslash e$ which contains $i$.
But this is automatically a basis of $M$ that contains $i$ but doesn’t contain
$e$. This is a contradiction, and suffices for the proof. ∎

######  Theorem 4.33 (Degree 1 Socles).

Let $M=(E,\mathcal{I})$ be a matroid satisfying $\operatorname{rank}(M)\geq
3$. Let $S\subseteq E(M)$ be a subset with
$\operatorname{rank}(S)\leq\operatorname{rank}(M)-2$. Then

$\\{\xi\in A_{f}^{1}:\xi(\partial_{i}f)=0\text{ for }i\in E\backslash
S\\}=\\{0\\}.$

###### Proof.

Let $r=\operatorname{rank}(M)\geq 3$. We want to prove that if a linear form
$\xi=\sum_{e\in E}c_{e}\cdot\partial_{e}$ satisfies $\xi(\partial_{e}f_{M})=0$
for all $e\in E\backslash S$, then we have $\xi(f_{M})=0$. For $i\in
E\backslash S$, we have

$0=\xi(\partial_{i}f_{M})=\sum_{e\in
E}c_{e}\partial_{e}\partial_{i}f_{M}=\sum_{e\in
E}c_{e}\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{r-2}(M)\\\
\alpha\cup\\{e,i\\}\in\mathcal{I}_{r}(M)\end{subarray}}x^{\alpha}=\sum_{\alpha\in\mathcal{I}_{r-2}(M)}\left(\sum_{\begin{subarray}{c}e\in
E\\\
\alpha\cup\\{i,e\\}\in\mathcal{I}_{r}(M)\end{subarray}}c_{e}\right)x^{\alpha}.$
(4.6)

By setting all of the coefficients of the right hand side of Equation 4.6, we
have that

$\sum_{\begin{subarray}{c}e\in E\\\
\alpha\cup\\{i,e\\}\in\mathcal{I}_{r}(M)\end{subarray}}c_{e}=0\quad\text{for
all $\alpha\in\mathcal{I}_{r-2}(M)$ and $i\in E\backslash S$}.$

For any $\beta\in\mathcal{I}_{r-1}(M)$, we know that $\beta\not\subseteq S$
since $\operatorname{rank}(\beta)=r-1>\operatorname{rank}(S)$. There exists
some $i\in\beta\backslash S$. Thus, we can write $\beta=\alpha\cup\\{i\\}$
where $\alpha\in\mathcal{I}_{r-2}(M)$ and $i\in E\backslash S$. This proves
that for any $\beta\in\mathcal{I}_{r-1}(M)$, we have

$\sum_{\begin{subarray}{c}e\in E\\\
\beta\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}c_{e}=\sum_{\begin{subarray}{c}e\in
E\\\ \alpha\cup\\{i,e\\}\in\mathcal{I}_{r}\end{subarray}}c_{e}=0.$

Finally, we have that

$\xi(f_{M})=\sum_{e\in E}c_{e}\partial_{e}f_{M}=\sum_{e\in
E}c_{e}\sum_{\begin{subarray}{c}\beta\in\mathcal{I}_{r-1}\\\
\beta\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}x^{\beta}=\sum_{\beta\in\mathcal{I}_{r-1}}\left(\sum_{\begin{subarray}{c}e\in
E\\\
\beta\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}c_{e}\right)x^{\beta}=0.$

This suffices for the proof. ∎

######  Theorem 4.34 (Higher Degree Socles).

Let $M=(E,\mathcal{I})$ be a matroid. Let $S\subseteq E$ be a subset such that
$\operatorname{rank}(S)\leq\operatorname{rank}(M)-k-1$. Then,

$\\{\xi\in A^{k}(M):\xi(\partial_{e}f_{M})=0\text{ for all }e\in E\backslash
S\\}=\\{0\\}.$

###### Proof.

Any $\xi\in A^{k}(M)$ can be written as
$\xi=\sum_{\alpha\in\mathcal{I}_{k}}c_{\alpha}\partial^{\alpha}$. For any
$e\in E\backslash S$, we have

$\displaystyle
0=\xi\partial_{i}f_{M}=\sum_{\alpha\in\mathcal{I}_{k}}c_{\alpha}\partial_{e}\partial^{\alpha}f_{M}=\sum_{\alpha\in\mathcal{I}_{k}}c_{\alpha}\sum_{\begin{subarray}{c}\gamma\in\mathcal{I}_{r-k-1}\\\
\gamma\cup\alpha\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}x^{\gamma}=\sum_{\gamma\in\mathcal{I}_{r-k-1}}\left(\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}\\\
\gamma\cup\alpha\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}c_{\alpha}\right)x^{\gamma}.$

This implies that for all $\gamma\in\mathcal{I}_{r-k-1}$ and $e\in E\backslash
S$, we have

$\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}\\\
\gamma\cup\alpha\cup\\{e\\}\in\mathcal{I}_{r}\end{subarray}}c_{\alpha}=0\quad\text{for
all $\gamma\in\mathcal{I}_{r-k-1}$ and $e\in E\backslash S$.}$

Let $\beta\in\mathcal{I}_{r-k}$ be an arbitrary independent set of rank $r-k$.
Note that $\operatorname{rank}(\beta)=r-k>\operatorname{rank}(S)$. Thus, we
cannot have $\beta\subseteq S$. This implies that we can find
$e\in\beta\backslash S$ such that $\beta=\alpha\cup\\{e\\}$ for $e\in
E\backslash S$. Thus,

$\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}\\\
\beta\cup\alpha\in\mathcal{I}_{r}\end{subarray}}c_{\alpha}=0\quad\text{for all
$\beta\in\mathcal{I}_{r-k}$}.$

Then, we have

$\xi(f_{M})=\sum_{\alpha\in\mathcal{I}_{k}}c_{\alpha}\partial^{\alpha}f_{M}=\sum_{\alpha\in\mathcal{I}_{k}}c_{\alpha}\sum_{\begin{subarray}{c}\beta\in\mathcal{I}_{r-k}\\\
\beta\cup\alpha\in\mathcal{I}_{r}\end{subarray}}x^{\beta}=\sum_{\beta\in\mathcal{I}_{r-k}}\left(\sum_{\begin{subarray}{c}\alpha\in\mathcal{I}_{k}\\\
\beta\cup\alpha\in\mathcal{I}_{r}\end{subarray}}c_{\alpha}\right)x^{\beta}=0,$

which suffices for the proof. ∎

We are now ready to state our main theorem about $\mathsf{HRR}_{1}$ on the
facets of $\mathbb{R}_{\geq 0}^{E}$. We study the properties of
$\operatorname{A}(M)$ on the relative interiors of the facets $H_{e}$ whenever
$e$ is not a coloop.

######  Theorem 4.35.

Let $M=(E,\mathcal{I})$ be a matroid which satisfies
$\operatorname{rank}(M)\geq 2$. For any $e\in E(M)$, the basis generating
polynomial $f_{M}$ satisfies $\mathsf{HRR}_{1}$ on
$\operatorname{relint}(H_{e})$ if and only if $e$ is not a co-loop.

###### Proof.

Without loss of generality, let $M$ be a matroid on the set $[n]$ and let
$e=1$. Then, we want to prove that whenever $a_{2},\ldots,a_{n}>0$, the ring
$A(M)$ satisfies $\mathsf{HRR}_{1}$ on $a=(0,a_{2},\ldots,a_{n})$ if and only
if $e$ is not a co-loop. We first prove that if $1$ is not a co-loop, then
$A(M)$ satisfies $\mathsf{HRR}_{1}$. To prove this, we induct on the rank of
$M$. For the base case $\operatorname{rank}(M)=2$, Corollary 4.25 implies that
it suffices to prove that $A(\widetilde{M})$ satisfies $\mathsf{HRR}_{1}$ on
$\Phi(l_{a})$. The only simple matroid of rank $2$ is the uniform matroid of
rank $2$. From Corollary 4.29, it suffices to check that the signature of the
Hessian is $(+,-,\ldots,-)$. But the Hessian of a rank $2$ simple matroid at
every point is the same matrix. If the matroid is on $n$ elements, the Hessian
matrix is

$A(K_{n})=\begin{bmatrix}0&1&1&\ldots&1\\\ 1&0&1&\ldots&1\\\ 1&1&0&\ldots&1\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ 1&1&1&\ldots&0\end{bmatrix}.$

This happens to be the adjancency matrix of a complete graph. From Proposition
1.5 in [58], this matrix has an eigenvalue of $-1$ with multiplicity $n-1$ and
an eigenvalue of $n-1$ with multiplicity $1$. Hence, its signature is
$(+,-,\ldots,-)$ which proves the base case. Now suppose that the claim is
true for all matroids of rank less than $r$. Let $M$ be a matroid of rank $r$.
We want to prove that the claim is true for $M$. By the same reasoning as in
the base case, we can assume that our matroid is simple. In the
simplification, the codimension of our $1$-form will not increase. For all
$i\in E\backslash\\{1\\}$, we know that $e$ is not a co-loop of $M/i$ from
Lemma 4.32. Thus, the simplified $l_{a}$ will either be all positive (in which
case the claim is known to be true) or $l_{a}$ will be all positive except
possibly at $e$. Since $e$ is known not to be a coloop of $M/i$, the
simplification of $l_{a}$ will be a $1$-form which satisfies the hypothesis in
the inductive hypothesis. Since $M$ is simple, we know that
$\operatorname{rank}(M/i)=\operatorname{rank}(M)-1<\operatorname{rank}(M)$.
Hence, from the inductive hypothesis, we know that
$A(M/i)=A_{\partial_{i}f_{M}}^{\bullet}$ satisfies $\mathsf{HRR}_{1}$ for
$l_{a}$.

Now, we have enough information to directly prove that $A(M)$ satisfies
$\mathsf{HRR}_{1}$ with respect to $l_{a}$. From Lemma 4.16, it suffices to
prove that $A(M)$ satisfies $\mathsf{HL}_{1}$ with respect to $l_{a}$. Since
$\dim_{\mathbb{R}}A^{1}(M)=\dim_{\mathbb{R}}A^{r-1}(M)$ from properties of
Poincaré Duality algebras, it suffices to prove that the Lefschetz operator
$L_{l_{a}}^{1}:A^{1}(M)\to A^{r-1}(M)$ is injective. Let $\Xi\in A^{1}(M)$ be
the kernel of $L_{l_{a}}^{1}$. Note that this is equivalent to requiring $\Xi
l_{a}^{r-2}=0$ in $A(M)$. We want to prove that $\Xi=0$ in $A^{1}(M)$. Since
$\Xi l_{a}^{r-2}=0$ in $A(M)$, we have

$0=-Q_{l_{a}}^{1}(\Xi,\Xi)=\deg_{M}(\Xi^{2}\cdot
l_{a}^{r-2})=\sum_{i=2}^{n}a_{i}\deg_{M}(\Xi^{2}\cdot
l_{a}^{r-3}\cdot\partial_{i}).$

Note that

$\deg_{M}(\Xi^{2}\cdot l_{a}^{r-3}\cdot\partial_{i})=(\Xi^{2}\cdot
l_{a}^{r-3})(\partial_{i}f_{M})=\deg_{M/i}(\Xi^{2}\cdot
l_{a}^{r-3})=-Q_{M/i}(\Xi,\Xi).$

where $Q_{M/i}$ is the Hodge-Riemann form of degree $1$ with respect to $l$
associated with $A(M/i)$. Thus,

$\sum_{i=2}^{n}a_{i}Q_{M/i}(\Xi,\Xi)=0.$

In $A(M/i)$, the linear form $\Xi$ is in the primitive subspace. Since we know
that $A(M/i)$ satisfies $\mathsf{HRR}_{1}$ with respect to $l$, we know that
the Hodge-Riemann form $Q_{M/i}$ is negative-definite on $\mathbb{R}\Xi$.
Since $a_{i}>0$ for $i=2,\ldots,n$, this implies that $Q_{M/i}(\Xi,\Xi)=0$ for
all such $i$. Hence $\Xi=0$ in $A(M/i)$ for all $i\in[2,n]$. In terms of
polynomials, this means that $\Xi(\partial_{i}f_{M})=0$ for all $i\in[2,n]$.
From Theorem 4.33, this implies that $\Xi(f_{M})=0$. Thus, $A(M)$ satisfies
$\mathsf{HRR}_{1}$ with respect to $l$.

For the other direction, we will prove that if $e$ is a co-loop of $M$, then
$f_{M}$ does not satisfy $\mathsf{HRR}_{1}$ on $\operatorname{relint}(H_{e})$.
Without loss of generality, we can suppose $E(M)=[n]$ and $e=1$. The linear
differential can be written as $l_{a}$ where $a=(0,a_{2},\ldots,a_{n})$ for
$a_{2},\ldots,a_{n}\geq 0$. It suffices to consider the case when $M$ is
simple because the coefficient of $\partial_{1}$ in the simplification of
$l_{a}$ will remain $0$. This is because co-loops have no parallel elements.
In the simple case, the bottom $(n-1)\times(n-1)$ sub-matrix of
$\operatorname{Hess}_{f_{M}}$ will be entirely $0$. This means that the Hodge-
Riemann form will singular and $f_{M}$ cannot satisfy $\mathsf{HRR}_{1}$ on
$\operatorname{relint}(H_{e})$. This suffices for the proof. ∎

With a same proof, we also have Theorem 4.36.

######  Theorem 4.36.

Let $M=(E,\mathcal{I})$ be a matroid which satisfies
$\operatorname{rank}(M)\geq 2$. For any $S\subseteq E$, if
$\operatorname{rank}(S)\leq\operatorname{rank}(M)-2$ and $S$ contains no co-
loops, then $f_{M}$ satisfies $\mathsf{HRR}_{1}$ on
$\operatorname{relint}(H_{S})$.

Theorem 4.36 in tandem with Theorem 4.31 immediately gives a somewhat random
computational result that under the hypothesis in Theorem 4.36, we have that

$\nabla f_{M/e}^{T}\cdot\operatorname{Hess}_{f_{M\backslash
e}}^{-1}\cdot\nabla f_{M/e}|_{x=a}\neq 0.$

Using our result about high degree socles in Theorem 4.34, we can prove an
inductive procedure for local $\mathsf{HRR}_{k}$ in Theorem 4.37 for points
$a\in\mathbb{R}_{\geq 0}^{E}$ whose support is large enough.

######  Theorem 4.37.

Let $M=(E,\mathcal{I})$ be a matroid and let $S\subseteq E$ be a subset such
that $\operatorname{rank}(S)\leq\operatorname{rank}(M)-k-1$. Let
$l:=\sum_{i\in E\backslash S}a_{i}\cdot e_{i}$ where $a_{i}>0$. If
$\partial_{e}f_{M}$ is $0$ or satisfies $\mathsf{HRR}_{k}$ and with respect to
$l$ for all $e\in E\backslash S$, then $A(M)$ satisfies $\mathsf{HL}_{k}$ with
respect to $l$.

###### Proof.

Note that $\partial_{e}f=0$ if and only if $e$ is a loop. In this case, the
variable $x_{e}$ doesn’t appear in the basis generating polynomial. Hence,
without loss of generality, we can suppose that $M$ is loopless and
$\partial_{e}f\neq 0$ for all $e\in E$. Since $A(M)$ is a Poincaré duality
algebra, it suffices to prove for $\Xi\in A^{k}(M)$ that if $\Xi l^{d-2k}=0$
in $A^{r-k}(M)$, then $\Xi=0$ in $A^{k}(M)$. We know that $\Xi$ is in the
primitive subspaces of $\partial_{e}f$ for all $e\in E$ and we can compute the
formula

$0=Q^{k}(\Xi,\Xi)=\sum_{i\in E\backslash S}a_{i}Q_{\partial_{i}f}(\Xi,\Xi).$

From the positive definiteness on the primitive subspaces, we have $\Xi=0$ in
$A^{k}_{\partial_{i}f}$ for all $i\in E\backslash S$. From Theorem 4.34, we
get $\Xi=0$ in $A^{k}(M)$. This suffices for the proof. ∎

From Theorem 4.37, it is enough to show that if $\operatorname{A}(M)$
satisfies $\mathsf{HRR}_{i}$ for $1\leq i\leq k-1$ and $\mathsf{HL}_{i}$ for
$1\leq i\leq k$, then it satisfies $\mathsf{HRR}_{k}$. Under these conditions,
satisfying $\mathsf{HRR}_{k}$ is equivalent to a statement about the net
signature of the Hodge-Riemann form. We define the notion of net signature in
Definition 4.38.

######  Definition 4.38.

Let $B:V\times V\to k$ be a symmetric bilinear form on a finite dimensional
vector space $V$. Suppose that the signature of $B$ has $n_{+}$ positive
eigenvalues and $n_{-}$ negative eigenvalues. Then, we define the net
signature to be $\sigma(B)=n_{+}-n_{-}$.

When our bilinear form $B:V\times V\to k$ is non-degenerate, then the net
signature $\sigma(B)$ determines the exact signature of the form. Indeed, in
the non-degenerate case, we have $n_{+}-n_{-}=\sigma(B)$ and $n_{+}+n_{-}=\dim
V$.

######  Lemma 4.39.

Let $A(M)$ satisfies $\mathsf{HRR}_{i}$ and $\mathsf{HL}_{i}$ with respect to
$l$ for $1\leq i\leq k$, then, we have

$\sigma\left((-1)^{k}Q_{l}^{k}\right)=\sum_{i=0}^{k}(-1)^{i}(\dim
A^{i}(M)-\dim A^{i-1}(M)).$

###### Proof.

We induct on $k$. For the base case, we have $k=1$ and the claim follows from
Proposition 4.15. Suppose that the claim holds for $k-1$. Consider the
composition of maps given by the following commutative diagram:

${{A^{k-1}(M)}}$${{A^{k}(M)}}$${{A^{d-k}(M)}}$${{A^{d-k+1}(M)}}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\times l_{a}^{d-2k}}$$\scriptstyle{\times
l_{a}}$$\scriptstyle{\psi_{a}}$ (4.7)

This diagram exhibits an isomorphism between $A^{k-1}(M)$ and $A^{d-k+1}(M)$
from $\mathsf{HL}_{k-1}$. This implies that we can decompose $A^{k}(M)$ into

$A^{k}(M)=l\cdot A^{k-1}(M)\oplus\ker\psi_{a}$

where the two summands in the direct sum are orthogonal with respect to the
Hodge-Riemann form $Q_{l}^{k}$. Let $u_{1},\ldots,u_{m}\in A^{k-1}(M)$ be a
basis for $A^{k-1}(M)$. Then, $l\cdot u_{1},\ldots,l\cdot u_{m}$ is a basis
for $A^{k}(M)$ and

$(-1)^{k}Q_{l}^{k}(l\cdot u_{i},l\cdot
u_{j})=(-1)^{k-1}Q_{l}^{k-1}(u_{i},u_{j}).$

Thus, the signature of $Q_{l}^{k}$ on $l\cdot A^{k-1}(M)$ should be the
negative of the signature of $Q_{l}^{k-1}$ on $A^{k-1}(M)$. Since $A(M)$
satisfies $\mathsf{HRR}_{k}$, we know that the signature of $Q_{l}^{k}$ on
$\ker\psi$ is $\dim\ker\psi$. This gives us the formula

$\displaystyle\sigma((-1)^{k}Q_{l_{a}}^{k})$
$\displaystyle=\sigma\left((-1)^{k-1}Q_{l_{a}}^{k-1}\right)+(-1)^{k}(\dim
A^{k}(M)-\dim A^{k-1}(M))$ $\displaystyle=\sum_{i=0}^{k}(-1)^{i}(\dim
A^{i}(M)-\dim A^{i-1}(M))$

from the inductive hypothesis. This suffices for the proof. ∎

######  Lemma 4.40.

Let $k\geq 2$ and $A(M)$ satisfies $\mathsf{HRR}_{i}$ and $\mathsf{HL}_{i}$
with respect to $l=l_{a}$ for some $a\in\mathbb{R}^{n}$ for all $1\leq i\leq
k-1$. Then

  1. (a)

$A(M)$ satisfies $\mathsf{HL}_{k}$ with respect to $l$ if and only if
$Q_{l}^{k}$ is non-degenerate on $A^{k}(M)$.

  2. (b)

Suppose that $\mathsf{HL}_{k}$ is satisfied. Then $A(M)$ satisfies
$\mathsf{HRR}_{k}$ with respect to $l$ if and only if

$\sigma((-1)^{k}Q_{l}^{k})=\sum_{i=0}^{k}(-1)^{i}(\dim A^{i}(M)-\dim
A^{i-1}(M)).$

###### Proof.

The argument for (a) is exactly the same as the argument for (a) in
Proposition 4.15. For part (b), consider the composition in Equation 4.7.
Recall from the proof of Lemma 4.39, we have that

$\sigma((-1)^{k}Q_{l_{a}}^{k})=\sigma\left((-1)^{k-1}Q_{l_{a}}^{k-1}\right)+\sigma\left((-1)^{k}Q_{l_{a}}^{k}|_{\ker\psi_{a}}\right).$

Since $A(M)$ satisfies $\mathsf{HRR}_{k}$ if and only if $Q_{l_{a}}^{k}$ is
positive definition on $\ker\psi_{a}$, we know that $A(M)$ satisfies
$\mathsf{HRR}_{k}$ if and only if $\sigma(Q_{l_{a}}^{k})=\dim\ker\psi_{a}=\dim
A^{k}(M)-\dim A^{k-1}(M)$. Then Lemma 4.39 completes the proof. ∎

In the situation where we know that $\operatorname{A}(M)$ satisfies
$\mathsf{HRR}_{i}$ for $1\leq i\leq k-1$ and $\mathsf{HL}_{i}$ for $1\leq
i\leq k$, we already know that $\mathsf{HRR}_{k}$ is equivalent to a statement
about the signature of the Hodge Riemann form. Note that as a function of the
point $a\in\mathbb{R}^{E}$ in $l_{a}$, the Hodge-Riemann form is continuous.
In particular, the eigenvalues of the Hodge-Riemann form are continuous
functions. This property will allow us to prove Lemma 4.41. This result tells
us that it is enough to prove that $\mathsf{HRR}_{k}$ and $\mathsf{HL}_{k}$
holds on at least one boundary point.

######  Lemma 4.41.

Let $\Omega\subseteq\mathbb{R}^{n}$ be a subset such that for all
$x\in\Omega$, $A(M)$ satisfies $\mathsf{HRR}_{i}$ for $i\leq k-1$ and
$\mathsf{HL}_{i}$ for $i\leq k$. Suppose that there is a continuous path
$\gamma:[0,1]\to\mathbb{R}^{n}$ with image $\varphi([0,1])\subseteq\Omega$,
suct that $A(M)$ satisfies $\mathsf{HRR}_{k}$ with respect to $l_{\gamma(0)}$.
Then $A(M)$ satisfies $\mathsf{HRR}_{k}$ with respect to $l_{\gamma(1)}$.

###### Proof.

Let $\lambda_{i}(a)$ be the $i$th largest eigenvalue of $Q_{l_{a}}^{k}$ as a
function in $a$ for $1\leq i\leq\dim A^{k}(M)$. Then for all $1\leq i\leq\dim
A^{k}(M)$, the $i$th eigenvalue $\lambda_{i}(a)$ is a continuous function in
$a$. Along the path $\gamma$, we know that $Q_{l_{a}}^{k}$ is non-degenerate.
Hence, none of the functions $\lambda_{i}(a)$ cross zero. This implies that on
the path, the signature $\sigma\left((-1)^{k}Q_{l_{a}}^{k}\right)$ remains
constant. From Lemma 4.40(b), this completes the proof. ∎

###  4.6 Future Work

As noted in the thesis, the Conjecture 4.28 was recently to be found
erroneous. However, we will still describe our thought process in our attempt
to prove it. We first proved Theorem 4.36. This describes the exact conditions
needed for $\operatorname{A}(M)$ to satisfy $\mathsf{HRR}_{1}$ on the faces of
the positive orthant. Theorem 4.37 tells us that for general $k$, we have a
mechanism from going from local $\mathsf{HRR}_{k}$ to $\mathsf{HL}_{k}$ on the
positive orthant and possibly its boundary depending on some dimension
conditions. This means that to prove the general conjecture, it suffices to
prove that we can go from $\mathsf{HRR}_{i}$ for $1\leq i\leq k-1$ and
$\mathsf{HL}_{i}$ for $1\leq i\leq k$ to $\mathsf{HRR}_{k}$. In Lemma 4.41, we
further reasoned that given $\mathsf{HRR}_{i}$ for $1\leq i\leq k-1$ and
$\mathsf{HL}_{i}$ for $1\leq i\leq k$ as the inductive hypothesis, it suffices
to prove that $\mathsf{HRR}_{k}$ is satisfied on a suitable boundary point.
The reason why we might want to pick a boundary point even though it may seem
more accessible is because we may hope for a semi-small decomposition for
$\operatorname{A}(M)$ as in Theorem 4.4 and Theorem 4.6. Indeed, if we were to
have such a decomposition, we could reduce $\mathsf{HRR}_{k}$ on the boundary
for $\operatorname{A}(M)$ to $\mathsf{HRR}_{k}$ for
$\operatorname{A}(M\backslash i)$. The conjecture would then follow from a
more elaborate induction argument. Before we can contemplate such an argument,
there are a few questions we must answer.

######  Question 1.

Does there exist a graded ring homomorphism
$\theta:\operatorname{A}(M\backslash i)\to\operatorname{A}(M)$ for every $i\in
E$?

In order to have a decomposition from which we can extract inductive
information about the Hodge structures, we must write $\operatorname{A}(M)$ as
the direct sum of $\operatorname{A}(M\backslash i)$ submodules. Even before
this, we would require that $\operatorname{A}(M)$ be a
$\operatorname{A}(M\backslash i)$-module. Note that there is a well-defined
surjection $\Theta_{M}:\operatorname{H}(M)\to\operatorname{A}(M)$ between the
graded Möbius algebra and the ring $\operatorname{A}(M)$ defined by

$\Theta_{M}(y_{F})=x^{I},\quad\text{ where $I$ is a basis of $F$}.$

A reasonable guess for an injection $\operatorname{A}(M\backslash
i)\to\operatorname{A}(M)$ would be the map which makes the diagram in Equation
4.8 commute.

${{\operatorname{H}(M\backslash
i)}}$${{\operatorname{H}(M)}}$${{\operatorname{A}(M\backslash
i)}}$${{\operatorname{A}(M)}}$ (4.8)

This map is forced to send the equivalence class of
$\operatorname{A}(M\backslash i)$ represented by a polynomial $f$ to the
equivalence class of $\operatorname{A}(M)$ represented by the same polynomial.
Unfortunately, it is not clear that this map is well-defined. In order to be
well-defined, Question 2 must have an affirmative answer.

######  Question 2.

Let $M$ be a matroid on the ground set $[n]$. Let
$\xi\in\mathbb{R}[\partial_{1},\ldots,\partial_{n-1}]$ such that
$\xi(f_{M\backslash n})=0$. Is it true that $\xi(f_{M})=0$?

In the case where $n$ is a coloop, the answer to Question 2 is Yes. When $n$
is not a coloop, we can write $f_{M}=x_{n}f_{M/n}+f_{M\backslash n}$. In this
case, Question 2 becomes equivalent to Question 3.

######  Question 3.

Let $M$ be a matroid and suppose that $e\in E(M)$ is not a coloop. Is it the
case that $\operatorname{Ann}(f_{M\backslash
e})\subseteq\operatorname{Ann}(f_{M/e})$?

At this time, we have not found a counter-example to Question 3. Note that the
operation $M\backslash e\to M/e$ is called a matroid quotient. Assuming that
the answers to all of these questions are Yes, then a decomposition of
$\operatorname{A}(M)$ may possibly descend from a decomposition of
$\operatorname{H}(M)$. Even if the mathematics doesn’t work in this way, the
question of having a semi-small decomposition of $\operatorname{H}(M)$ is an
interesting question. We know that $\operatorname{H}(M)$ is a
$\operatorname{H}(M\backslash i)$ module. From the Krull-Schmidt theorem,
there is some unique $\operatorname{H}(M\backslash i)$-module decomposition of
the form

$\operatorname{H}(M)=S_{1}\oplus\ldots\oplus S_{m}$

where the $S_{i}$’s are indecomposable $\operatorname{H}(M\backslash
i)$-modules. If we look at the $0$-degree parts of these modules, we know that
$H^{0}(M)\cong\mathbb{R}$. Thus, all but one of the $S_{i}$’s must have
trivial $0$-degree parts. This means that $1\in S_{i}$ for some $i\in[m]$.
Since they are $\operatorname{H}(M\backslash i)$-modules, this means that
$\operatorname{H}(M\backslash i)\subseteq S_{i}$ for some $i\in[m]$. This
gives us some idea on how a semi-small decomposition $\operatorname{H}(M)$
could be constructed. However, we discovered that $\operatorname{H}(M)$ does
not necessary have a semi-small decomposition. Indeed, consider the matroid
that comes from the affine configuration of $\\{A,B,C,D,E\\}$ in Figure 4.1.
One can calculate that as $\operatorname{H}(M\backslash A)$ modules, the ring
$\operatorname{H}(M)$ is indecomposable. This does not bode well for the
possibility for a general semi-small decomposition.

Figure 4.1: $\operatorname{H}(M)$ is indecomposable

However, this may be a result of picking a bad point to delete. Perhaps, there
always exists a point that we can delete to give a nice decomposition. Another
possibility is that there may be a decomposition in low dimensions
($\leq\frac{d}{2}$). For example, one conjecture is that of Question 4, which
asks about the equivalence of $\operatorname{A}(M)$ and $\operatorname{H}(M)$
in small dimensions.

######  Question 4.

Is it true that $\operatorname{H}^{k}(M)\to\operatorname{A}^{k}(M)$ is an
isomorphism when $k\leq\frac{d}{2}$?

For any graded ring $R$ of top dimension $d$, we let $D^{k}R$ be the graded
ring of dimension $k$ where we annihilate all elements of degree greater than
$k$. Question 5 asks if there is a possibility of a semi-small decomposition
of $\operatorname{H}(M)$ for a small enough $k$?

######  Question 5.

Is $D^{k}\operatorname{H}(M\backslash i)$ a summand in the Krull-Schmidt
decomposition of $D^{k}\operatorname{H}(M)$ for $k\leq\frac{d}{2}$?

However, much of this strategy will not work because of a counter-example to
Conjecture 4.28 that was provided to us by Connor Simpson. This counter-
example also gives a negative answer to Question 4. Let $G=K_{2,3}$ be the
complete bipartite graph on $2+3$ vertices. Let $M=M(G)$ be the graphic
matroid of rank $4$ with $G$ as the underlying graph. Then, $M$ has $15$ flats
of rank $2$. The pairing on $\operatorname{H}^{2}(M)$ has $6$ positive
eigenvalues, $6$ negative eigenvalues, and $3$ zero eigenvalues. The zero
eigenvalues implies that the answer of Question 4 is no. The negative
eigenvalues gives the contradiction to Conjecture 4.28. However, there is
still a possibility for Conjecture 4.27 to be true. If it were, this would
provide an example of a polynomial algebra for which Hard Lefschetz holds but
in general the Hodge-Riemann relations do not.

##  Chapter 5 Appendix

###  5.1 Brunn-Minkowski and the Base Case of the Alexandrov-Fenchel
Inequality

######  Proposition 5.1.

Let $K,L\subseteq\mathbb{R}^{2}$ be two convex bodies in the plane. Then
$\mathsf{V}_{2}(K,L)^{2}\geq\mathsf{V}(K,K)\mathsf{V}(L,L)$.

###### Proof.

From Theorem 2.17, we have
$\sqrt{\operatorname{Vol}_{2}(K+L)}\geq\sqrt{\operatorname{Vol}_{2}(A)}+\sqrt{\operatorname{Vol}_{2}(B)}$.
Squaring both sides, we get the equivalent inequality

$\operatorname{Vol}_{2}(K+L)-\operatorname{Vol}_{2}(K)-\operatorname{Vol}_{2}(L)\geq
2\sqrt{\operatorname{Vol}_{2}(K)\operatorname{Vol}_{2}(L)}=2\sqrt{\mathsf{V}_{2}(K,K)\mathsf{V}_{2}(L,L)}.$

From Theorem 1.53, the left hand side is equal to

$\operatorname{Vol}_{2}(K+L)-\operatorname{Vol}_{2}(K)-\operatorname{Vol}_{2}(L)=2\mathsf{V}_{2}(K,L).$

This suffices for the proof. ∎

###  5.2 Computations in the Kahn-Saks Inequality

Let $(P,\leq)$ be a finite poset and fix two elements $x,y\in P$. For the sake
of simplicity, we can assume that $x\leq y$. In the proof of Theorem 3.7, we
defined special cross-sections of the order polytope for every
$\lambda\in[0,1]$ given by

$K_{\lambda}:=\\{t\in\mathcal{O}_{P}:t_{y}-t_{x}=\lambda\\}.$

We give a proof of Lemma 5.2. This result is important in the proof of Theorem
3.7 as it gives a way to interpret the volume of $K_{\lambda}$ in terms of
mixed volumes. The proof of Lemma 5.2 was stated in [34] but was not proven.
We include the proof for the sake of completeness.

######  Lemma 5.2.

For $\lambda\in[0,1]$ we have that $K_{\lambda}=(1-\lambda)K_{0}+\lambda
K_{1}$.

###### Proof.

For all $\lambda\in\mathbb{R}$, we can define the hyperplanes
$H_{\lambda}=\\{t\in\mathbb{R}^{n}:t_{y}-t_{x}=\lambda\\}$. Then
$K_{\lambda}=H_{\lambda}\cap\mathcal{O}_{P}$. It is a simple computation to
prove that $H_{\lambda}=(1-\lambda)H_{0}+\lambda H_{1}$. For any linear
extension $\sigma\in e(P)$, we define the polytopes:

$\displaystyle\Delta_{\sigma}$ $\displaystyle:=\\{0\leq
t_{\sigma^{-1}(1)}\leq\ldots\leq t_{\sigma^{-1}(n)}\leq
1\\}\subseteq\mathcal{O}_{P}$ $\displaystyle\Delta_{\sigma}(\lambda)$
$\displaystyle:=\\{t\in\Delta_{\sigma}:t_{y}-t_{x}=\lambda\\}=K_{\lambda}\cap\Delta_{\sigma}=H_{\lambda}\cap\Delta_{\sigma}.$

Then, we can decompose $K_{\lambda}$ into

$K_{\lambda}=H_{\lambda}\cap\mathcal{O}_{P}=H_{\lambda}\cap\bigcup_{\sigma\in
e(P)}\Delta_{e}=\bigcup_{\sigma\in e(P)}\Delta_{\sigma}(\lambda).$

where the polytopes in the unions are disjoint up to a set of measure zero.
Now, note that we have

$\displaystyle(1-\lambda)K_{0}+\lambda K_{1}$
$\displaystyle\subseteq(1-\lambda)\mathcal{O}_{P}+\lambda\mathcal{O}_{P}=\mathcal{O}_{P}$
$\displaystyle(1-\lambda)K_{0}+\lambda K_{1}$
$\displaystyle\subseteq(1-\lambda)H_{0}+\lambda H_{1}=H_{\lambda}.$

Thus, we have that $(1-\lambda)K_{0}+\lambda K_{1}\subseteq\mathcal{O}_{P}\cap
H_{\lambda}=K_{\lambda}$. It suffices to prove the opposite inclusion. Let
$t\in K_{\lambda}$ be an arbitrary point. From our decomposition of
$K_{\lambda}$ into polytopes $\Delta_{\sigma}(\lambda)$, we know that there is
some linear extension $\sigma$ satisfying $\sigma(x)<\sigma(y)$ and
$t\in\Delta_{\sigma}(\lambda)$. The linear extension $\sigma$ induces a total
order $<_{\sigma}$ on $P$ defined by $z_{1}<_{\sigma}z_{2}$ if and only if
$\sigma(z_{1})<\sigma(z_{2})$. In particular, when
$\omega_{1}\geq_{\sigma}\omega_{2}$, we have that $t_{\omega_{1}}\geq
t_{\omega_{2}}$. Now, we can define two points $P_{0},P_{1}\in\mathbb{R}^{n}$
so that for all $\omega\in P$, we have

$\displaystyle(P_{0})_{\omega}$
$\displaystyle=\begin{cases}\frac{t_{\omega}}{1-\lambda}&\text{if
$\omega\leq_{\sigma}x$}\\\ \frac{t_{x}}{1-\lambda}&\text{if
$x<_{\sigma}\omega<_{\sigma}y$}\\\
\frac{t_{\omega}-\lambda}{1-\lambda}&\text{if
$\omega\geq_{\sigma}y$}.\end{cases}$ $\displaystyle(P_{1})_{\omega}$
$\displaystyle=\begin{cases}0&\text{if $\omega\leq_{\sigma}x$}\\\
\frac{t_{\omega}-t_{x}}{\lambda}&\text{if $x<_{\sigma}\omega<_{\sigma}y$}\\\
1&\text{if $\omega\geq y$}.\end{cases}$

Then $P_{0}\in K_{0}$, $P_{1}\in K_{1}$, and $t=(1-\lambda)P_{0}+\lambda
P_{1}$. Since $t$ was arbitrary, this completes the proof that
$K_{\lambda}=(1-\lambda)K_{0}+\lambda K_{1}$. ∎

######  Lemma 5.3.

For $i\in\\{1,\ldots,n-1\\}$, we have that
$N_{i}=(n-1)!\mathsf{V}_{n-1}(K_{0}[n-i],K_{1}[i-1])$.

###### Proof.

From Theorem 1.49 and Lemma 5.2, we have that

$\displaystyle\operatorname{Vol}_{n-1}(K_{\lambda})$
$\displaystyle=\operatorname{Vol}_{n-1}((1-\lambda)K_{0}+\lambda K_{1})$
$\displaystyle=\sum_{j=0}^{n-1}\binom{n-1}{j}\mathsf{V}_{n-1}(K_{0}[n-j-1],K_{1}[j])(1-\lambda)^{n-j-1}\lambda^{j}.$

Alternatively, we can compute the volume
$\operatorname{Vol}_{n-1}(K_{\lambda})$ as

$\displaystyle\operatorname{Vol}_{n-1}(K_{\lambda})$
$\displaystyle=\operatorname{Vol}_{n-1}\left(\bigcup_{\begin{subarray}{c}\sigma\in
e(P)\\\ \sigma(x)<\sigma(y)\end{subarray}}\Delta_{\sigma}(\lambda)\right)$
$\displaystyle=\sum_{\begin{subarray}{c}\sigma\in e(P)\\\
\sigma(x)<\sigma(y)\end{subarray}}\operatorname{Vol}_{n-1}(\Delta_{\sigma}(\lambda))$
$\displaystyle=\sum_{k=1}^{n-1}\sum_{\begin{subarray}{c}\sigma\in e(P)\\\
\sigma(y)-\sigma(x)=k\end{subarray}}\operatorname{Vol}_{n-1}(\Delta_{\sigma}(\lambda)).$

To compute $\operatorname{Vol}_{n-1}(\Delta_{\sigma}(\lambda))$, let
$P=\\{z_{1},\ldots,z_{n}\\}$ with $x=z_{i}$, $y=z_{j}$ with $\sigma(z_{k})=k$
for $k\in[n]$. In this notation, we have the explicit description of
$\Delta_{\sigma}(\lambda)$ as

$\Delta_{\sigma}(\lambda)=\\{t\in\mathbb{R}^{n}:0\leq t_{1}\leq\ldots\leq
t_{n}\leq 1\text{ and }t_{j}-t_{i}=\lambda\\}.$

This set of inequalities is equivalently decribed by the inequalities

$\displaystyle 0$ $\displaystyle\leq t_{i+1}-t_{i}\leq\ldots\leq
t_{j-1}-t_{i}\leq\lambda$ $\displaystyle 0$ $\displaystyle\leq
t_{1}\leq\ldots\leq t_{i-1}\leq t_{i}\leq t_{j+1}-\lambda\leq\ldots\leq
t_{n}-\lambda\leq 1-\lambda.$

Thus, the volume of $\Delta_{n}(\lambda)$ ends up being the product of two
simplices. Specifically, we have the equation

$\operatorname{Vol}_{n-1}(\Delta_{\sigma}(\lambda))=\frac{\lambda^{j-i-1}}{(j-i-1)!}\cdot\frac{(1-\lambda)^{n-(j-i)}}{(n-(j-i))!}.$

We can finish the computation with

$\displaystyle\operatorname{Vol}_{n-1}(K_{\lambda})$
$\displaystyle=\sum_{k=1}^{n-1}\sum_{\begin{subarray}{c}\sigma\in e(P)\\\
\sigma(y)-\sigma(x)=k\end{subarray}}\frac{\lambda^{k-1}}{(k-1)!}\cdot\frac{(1-\lambda)^{n-k}}{(n-k)!}$
$\displaystyle=\sum_{k=1}^{n-1}\binom{n-1}{k-1}\frac{N_{k}}{(n-1)!}\cdot\lambda^{k-1}(1-\lambda)^{n-k}.$

Thus, we have that

$\frac{N_{k}}{(n-1)!}=\mathsf{V}_{n-1}(K_{0}[n-k],K_{1}[k-1])\implies
N_{k}=(n-1)!\mathsf{V}_{n-1}(K_{0}[n-k],K_{1}[k-1]).$

This suffices for the proof of the Lemma. ∎

######  Lemma 5.4.

The affine hulls of $K_{0}$, $K_{1}$, and $K_{0}+K_{1}$ are given by

$\displaystyle\operatorname{aff}(K_{0})$
$\displaystyle=\mathbb{R}\left[\sum_{\omega\in P_{x\leq\cdot\leq
y}}e_{\omega}\right]\oplus\bigoplus_{\omega\in P\backslash P_{x\leq\cdot\leq
y}}\mathbb{R}[e_{\omega}]$ $\displaystyle\operatorname{aff}(K_{1})$
$\displaystyle=\sum_{\omega\in P_{\geq y}}e_{\omega}+\bigoplus_{\omega\in
P\backslash{P_{\leq x}\cup P_{\geq y}}}\mathbb{R}[e_{\omega}]$
$\displaystyle\operatorname{aff}(K_{0}+K_{1})$ $\displaystyle=\sum_{\omega\in
P_{\geq y}}e_{\omega}+\mathbb{R}\left[\sum_{\omega\in P_{x\leq\cdot\leq
y}}e_{\omega}\right]\oplus\bigoplus_{\omega\in
P\backslash\\{x,y\\}}\mathbb{R}[e_{\omega}]$

where $\mathbb{R}[v]$ denotes the linear span of the vector $v$. As an
immediate corollary, we have

$\displaystyle\dim K_{0}$ $\displaystyle=n-|P_{x<\cdot<y}|-1$
$\displaystyle\dim K_{1}$ $\displaystyle=n-|P_{<x}|-|P_{>y}|-2$
$\displaystyle\dim(K_{0}+K_{1})$ $\displaystyle=n-1.$

###### Proof.

From Proposition 5.6, there is a linear extension $f:P\to[n]$ such that
$f(y)-f(x)=|P_{x<\cdot<y}|+1$. Then, there are elements
$\alpha,\beta_{j},\gamma_{k}$ for $1\leq i\leq a$, $1\leq j\leq b$, and $1\leq
k\leq c$ such that

$f(\alpha_{1})<\ldots<f(\alpha_{a})<f(x)<f(\beta_{1})<\ldots<f(\beta_{b})<f(y)<f(\gamma_{1})<\ldots<f(\gamma_{c}).$

This allows us to compute

$\displaystyle
K_{0}\cap\Delta_{f}=\\{t\in[0,1]^{n}:t_{\alpha_{1}}\leq\ldots\leq
t_{\alpha_{a}}\leq t_{x}=\ldots=t_{y}\leq t_{\gamma_{1}}\leq\ldots\leq
t_{\gamma_{c}}\\}.$

In particular, by taking affine spans, we have

$\operatorname{aff}K_{0}\supseteq\operatorname{aff}(K_{0}\cap\Delta_{f})=\mathbb{R}\left[\sum_{\omega\in
P_{x\leq\cdot\leq y}}e_{\omega}\right]\oplus\bigoplus_{\omega\in P\backslash
P_{x\leq\cdot\leq y}}\mathbb{R}[e_{\omega}].$

To prove the other inclusion, let $v\in K_{0}$ be an arbitrary vector. Since
we are working in a subset of the order polytope, it must be the case that
$v_{x}\leq v_{\omega}\leq v_{y}$ for all $\omega\in P_{x\leq\cdot\leq y}$. But
since $v_{x}=v_{y}$, this means that $v_{\omega}=v_{x}=v_{y}$ are all
$\omega\in P_{x\leq\cdot\leq y}$. Thus

$v\in\mathbb{R}\left[\sum_{\omega\in P_{x\leq\cdot\leq
y}}e_{\omega}\right]\oplus\bigoplus_{\omega\in P\backslash P_{x\leq\cdot\leq
y}}\mathbb{R}[e_{\omega}].$

Since $v$ is arbitrary, we have proved the given formula for
$\operatorname{aff}K_{0}$.

To prove the formula for $K_{1}$, we first use Proposition 5.7 to construct a
linear extension $g:P\to[n]$ satisfying $g(x)=1+|P_{<x}$ and
$g(y)=n-|P_{>y}|$. Note that

$K_{1}=\\{t\in\mathcal{O}_{P}:t_{y}-t_{x}=1\\}=\\{t\in\mathcal{O}_{P}:t_{y}=1,t_{x}=0\\}.$

In particular, for any $v\in K_{1}$, for all $\omega\leq x$ and $\eta\geq y$
we must have $v_{\omega}=0$ and $v_{\eta}=1$. Thus,

$K_{1}\subseteq\sum_{\omega\in P_{\geq y}}e_{\omega}+\bigoplus_{\omega\in
P\backslash(P_{\leq x}\cup P_{\geq y})}\mathbb{R}[e_{\omega}].$

On the other hand, if we have $P_{\leq
x}=\\{\alpha_{1},\ldots,\alpha_{a},x\\}$ ordered by according to $g$ and
$P_{\geq y}=\\{y,\beta_{1},\ldots,\beta_{b}\\}$ ordered according to $g$, then

$K_{1}\cap\Delta_{g}=\\{t\in[0,1]^{n}:0=t_{\alpha_{1}}\leq\ldots\leq
t_{\alpha_{a}}\leq t_{x}\leq\ldots\leq
t_{y}=t_{\beta_{1}}=\ldots=t_{\beta_{b}}=1\\}.$

Taking the affine span, we have

$\operatorname{aff}K_{1}\supseteq\operatorname{aff}(K_{1}\cap\Delta_{g})=\sum_{\omega\in
P_{\geq y}}e_{\omega}+\bigoplus_{\omega\in P\backslash(P_{\leq x}\cup P_{\geq
y}}\mathbb{R}[e_{\omega}].$

This completes the proof for the given formula for $\operatorname{aff}K_{1}$.
The third formula follows from the fact that
$\operatorname{aff}(K+L)=\operatorname{aff}K+\operatorname{aff}L$ for any non-
empty sets $K$ and $L$. ∎

####  5.2.1 Modifications and linear extensions of posets

In this section, we will define certain linear extensions and linear extension
modifications to help with the analysis of the polytopes associated to the
Kahn-Saks sequence. In particular, it will also be useful to have examples of
extremal linear extensions which place $x$ and $y$ close to each or very far
from each other. It will also be useful to modify an existing linear extension
to make a covering relation adjacent in a linear extension with changing too
many terms in the original extension.

######  Proposition 5.5.

The poset $P$ as at least one linear extension.

###### Proof.

We induct on the size of $P$. If $|P|=1$, then the partial order is already a
total order. Now suppose the claim is true for posets of size $n-1$ and let
$P$ be a poset of size $n$. Let $m\in P$ be a maximal element. Let
$P^{\prime}$ be the poset which you get by deleting $m$. From the inductive
hypothesis, there is a linear extension $g:P^{\prime}\to[n]$. Then, the map
$f:P\to[n]$ defined by

$f(\omega)=\begin{cases}g(\omega)&\text{ if $\omega\neq m$},\\\ n&\text{ if
$\omega=m$}\end{cases}$

is a linear extension of $P$. ∎

######  Proposition 5.6.

There exists a linear extension $f:P\to[n]$ satisfying
$f(y)-f(x)=|P_{x<\cdot<y}|+1$.

###### Proof.

Let $\mathscr{L}$ denote the set of linear extensions of $P$. From Proposition
5.5, there exists a linear extension $f_{\mathsf{min}}\in\mathscr{L}$ which
minimizes $f(y)-f(x)$ out of all $f\in\mathscr{L}$. Since $x\leq y$, we know

$f_{\mathsf{min}}(y)-f_{\mathsf{min}}(x)\geq 1+|P_{x<\cdot<y}|.$

If $f_{\mathsf{min}}(y)-f_{\mathsf{min}}(x)=1+|P_{x<\cdot<y}|$, then we
proposition is proved. Otherwise, we can assume
$f_{\mathsf{min}}(y)-f_{\mathsf{min}}(x)\geq 2+|P_{x<\cdot<y}|$. Define the
set

$\displaystyle\mathsf{M}_{f_{\mathsf{min}}}$ $\displaystyle:=\\{z\in
P:f_{\mathsf{min}}(x)<f_{\mathsf{min}}(z)<f_{\mathsf{min}}(y)\\}$
$\displaystyle=f_{\mathsf{min}}^{-1}\left\\{(f_{\mathsf{min}}(x),f_{\mathsf{min}}(y))\right\\}$

which consist of the elements of the poset which appear between $x$ and $y$ in
the linear extension $f_{\mathsf{min}}$. From the assumption
$f_{\mathsf{min}}(y)-f_{\mathsf{min}}(x)\geq 2+|P_{x<\cdot<y}|$, we know there
is at least one element in $\mathsf{M}_{f_{\mathsf{min}}}$ which is not
comparable to both $x$ and $y$. This is because any element that is comparable
to both $x$ and $y$ must be in the set $P_{x<\cdot<y}$ which doesn’t have
enough elements to satisfy the inequality. Without loss of generality, suppose
that there exists an element which is not comparable to $x$. Let
$z\in\mathsf{M}_{f_{\mathsf{min}}}$ be such an element which minimizes
$f_{\mathsf{min}}(z)$. In other words, in the linear extension, it is the
leftmost element of this type. All elements between $x$ and $z$ in the linear
extension must be greater than $x$ from our choice of $z$. In particular,
those elements cannot be less than $z$ since that would imply the relation
$x<z$. This means that we can move $z$ to the immediate left of $x$ and still
have a working linear extension. But the resulting linear extension will have
a smaller value of $f(y)-f(x)$. This contradicts the minimality of
$f_{\mathsf{min}}$ and completes the proof of the proposition. ∎

######  Proposition 5.7.

There exists a linear extension $f:P\to[n]$ satisfying $f(x)=|P_{<x}|+1$ and
$f(y)=n-|P_{>y}|$.

###### Proof.

Consider any linear extensions of the three subposets: $P_{\leq x}$,
$P\backslash(P_{\leq x}\cup P_{\geq y})$, and $P_{\geq y}$. By appending the
extensions in this order, we get an extension of $P$ of the desired form. ∎

######  Proposition 5.8.

Let $f:P\to[n]$ be a linear extension and let $a,b\in P$ be elements such that
$a\lessdot b$. Then there is a linear extension $\tilde{f}:P\to[n]$ satisfying
$\tilde{f}(b)-\tilde{f}(a)=1$ and $\tilde{f}(z)=f(z)$ whenever $f(z)<f(a)$ or
$f(z)>f(b)$.

###### Proof.

Consider the sub-poset consisting of $a$, $b$, and the elements in between $a$
and $b$ in the linear extension $f$. By Proposition 5.6, there is a linear
extension of this sub-poset so that $a$ lies to the immediate left of $b$. By
replacing the portion of the extension $f$ containing the elements of this
sub-poset by this linear extension, we get the desired linear extension. ∎

######  Lemma 5.9.

Let $(P,\leq)$ be a poset and let $x,y\in P$ be two elements such that
$x\not\geq y$. Suppose that $x$ and $y$ are incomparable with respect to the
order $\leq$. Let $\leq_{*}$ be a binary relation defined by
$z_{1}\leq_{*}z_{2}$ if and only if $z_{1}\leq z_{2}$ or $z_{1}\leq x$ and
$z_{2}\geq y$. Then $\leq_{*}$ is a partial order on $P$.

###### Proof.

In the following, we separate the proofs of reflexivity, anti-symmetry, and
transitivity for the binary relation $\leq_{*}$. Let $a,b,c\in P$ be arbitrary
points in the poset.

  1. (i)

Since $a\leq a$, we must have $a\leq_{*}a$. This proves reflexivity of
$\leq_{*}$.

  2. (ii)

If $a\leq_{*}b$ and $b\leq_{*}a$, then there are four possibilities. The first
possibility is $a\leq b$ and $b\leq a$. In this case, we get $a=b$. The second
possibility is $a\leq x,y\leq b$ and $b\leq x,y\leq a$. In this case, $y\leq
a\leq x$ so $y\leq x$. But this contradicts our assumption that $x\not\geq y$.
The third possibility is $a\leq b$ and $b\leq x$, $y\leq a$. In this case, we
have $y\leq a\leq b\leq x$ which cannot happen for the same reason the second
possibility cannot happen. Similarly, the fourth possibility cannot occur.
Thus, in any case, $\leq_{*}$ satisfies the anti-symmetry property.

  3. (iii)

For the final property, suppose that $a\leq_{*}b$ and $b\leq_{*}c$. Again,
there are four possibilities. The first possibility is $a\leq b$ and $b\leq
c$. Then $a\leq c$ which implies $a\leq_{*}c$. The second possibility is
$a\leq b$, $b\leq x$, and $y\leq c$. In this case, $a\leq x$ and $y\leq c$.
Thus $a\leq_{*}c$ holds. The third possibility is $a\leq x$, $y\leq b$, and
$b\leq c$. In this case $a\leq x$ and $y\leq c$. Thus $a\leq_{*}c$ holds. The
fourth possibility is $a\leq x,y\leq b$ and $b\leq x,y\leq c$. But this
implies that $y\leq x$ which cannot happen. In all cases, $\leq_{*}$ satisfies
the transitivity property.

∎

######  Lemma 5.10.

Let $(P,\leq)$ be a poset and $x,y\in P$ be elements such that $x\not\geq y$.
Let $\leq_{*}$ be the partial order as defined in Lemma 5.9. Let $\\{N_{k}\\}$
be the Kahn-Saks sequence with respect to $\leq$ and let
$\\{\widetilde{N}_{k}\\}$ be the Kahn-Saks sequence with respect to
$\leq_{*}$. Then $N_{k}=\widetilde{N}_{k}$ for all $k$.

###### Proof.

We prove the stronger statement that a bijective map $f:P\to[n]$ is a linear
extension of $\leq$ satisfying $f(y)-f(x)=k$ if and only if it is a linear
extension of $\leq_{*}$ satisfying $f(y)-f(x)=k$. Clearly any linear extension
of $\leq_{*}$ is a linear extension of $\leq$. Now suppose that $f$ is a
linear extension of $\leq$ satisfying $f(y)-f(x)=k$. Suppose for the sake of
contradiction that it is not a linear extension of $\leq_{*}$. This means that
there are two elements $a,b\in P$ such that $a<_{*}b$ and $f(b)<f(a)$. Since
$a<_{*}b$, we either have $a<b$ or $a\leq x$ and $y\leq b$. In the first case,
we would have $f(a)<f(b)$ since $f$ is a linear extension of $\leq$. This
would be give a contradiction. For the second case, we would have $f(a)\leq
f(x)<f(y)\leq f(b)$, which is also a contradiction. This suffices for the
proof. ∎

Lemma 5.11 is a linear algebraic result which allows us skimp on calculating
the exact affine span of our faces.

######  Lemma 5.11.

Let $W_{1}$ and $W_{2}$ be finite dimensional affine spaces. Let
$V_{1}\subseteq W_{1}$ and $V_{2}\subseteq W_{2}$ be affine subspaces such
that $\dim V_{1}=\dim W_{1}-a$ and $\dim V_{2}=\dim W_{2}-b$ for some $a,b\geq
0$. Then $\dim(V_{1}+V_{2})\geq\dim(W_{1}+W_{2})-a-b$.

###### Proof.

After suitable translations, it suffices to prove the result when all our
spaces are vector spaces. Then, we have

$\displaystyle\dim(V_{1}+V_{2})$ $\displaystyle=\dim V_{1}+\dim
V_{2}-\dim(V_{1}\cap V_{2})$ $\displaystyle\geq\dim W_{1}+\dim
W_{2}-\dim(W_{1}\cap W_{2})-a-b$ $\displaystyle=\dim(W_{1}+W_{2})-a-b.$

This completes the proof of the lemma. ∎

Our main application of Lemma 5.11 is when $W_{1}$ and $W_{2}$ are the affine
spans of $K_{0}$ and $K_{1}$ while $V_{1}$ and $V_{2}$ are the affine spans of
faces of $K_{0}$ and $K_{1}$.

###  5.3 Stanley Basis Inequality for Graphic Matroids

######  Lemma 5.12.

Theorem 3.35 is true when $G=(V,E)$ is a connected graph on $|V|=3$ vertices.

###### Proof.

The only non-trivial case to check is for $n=3$. We can define
$r_{1}=e_{R}(v_{2},v_{3})$, $r_{2}=e_{R}(v_{1},v_{3})$, and
$r_{3}=e_{R}(v_{1},v_{2})$. We can then define $q_{1},q_{2}$, and $q_{3}$
similarly. From the same reasoning as in Theorem 3.35, we get that

$\displaystyle r_{1}$
$\displaystyle=\lambda_{1}q_{1},r_{2}=\lambda_{2}q_{2},r_{3}=\lambda_{3}q_{3}$

and

$\displaystyle r_{1}+r_{2}$
$\displaystyle=\lambda_{1}(q_{1}+q_{2})=\lambda_{2}(q_{1}+q_{2})$
$\displaystyle r_{2}+r_{3}$
$\displaystyle=\lambda_{2}(q_{2}+q_{3})=\lambda_{3}(q_{2}+q_{3})$
$\displaystyle r_{3}+r_{1}$
$\displaystyle=\lambda_{3}(q_{3}+q_{1})=\lambda_{1}(q_{3}+q_{1})$

for some constants $\lambda_{1},\lambda_{2},\lambda_{3}\geq 0$. All of these
equations imply that $\lambda_{1}=\lambda_{2}=\lambda_{3}$. This suffices for
the proof. ∎

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