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\begin{document}
\lecture{10}{Lionel Levine}{March 8, 2011}{Alex Zhai}
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\section{Review of incidence algebras}
Recall that for a poset $P$ and field $K$, we have defined the
incidence algebra $I(P)$ to be the set of functions mapping intervals
of $P$ to $K$. To simplify notation, for $\alpha \in I(P)$ and $x, y
\in P$ with $x \le y$, we will write $\alpha(x, y)$ for $\alpha([x,
y])$. For $\alpha, \beta \in I(P)$, we defined the product $\alpha
\beta$ by
\[ (\alpha \beta)(x, y) = \sum_{x \le z \le y} \alpha(x, z) \beta(z, y). \]
This can be interpreted as a matrix multiplication. Associate to each
$\alpha \in I(P)$ a $|P| \times |P|$ matrix $M(\alpha)$ with rows and
columns indexed by elements of $P$, in which the $xy$ entry is
$\alpha(x, y)$ if $x \le y$ and $0$ otherwise. Then, it is a routine
exercise to verify that $M(\alpha)M(\beta) = M(\alpha \beta)$.
From the matrix perspective, it is clear that the multiplication we
have defined is associative with identity element $1(x, y) =
\delta_{xy}$. Thus, the incidence algebra $I(P)$ has the structure of
a $K$-algebra, justifying its appellation. Furthermore, we may
naturally say that $\beta = \alpha^{-1}$ if $\alpha \beta = 1$ (or,
equivalently, $\beta \alpha = 1$). In more explicit terms, this
condition is
\[ \sum_{x \le z \le y} \alpha(x, z) \beta(z, y) = \delta_{xy} \]
for all $x, y \in P$ with $x \le y$. Rearranging the equation yields
\[ \beta(x, y) = \frac{1}{\alpha(x, x)} \left( \delta_{xy} - \sum_{x < z \le y} \alpha(x, z) \beta(z, y) \right), \]
which provides us a way of solving for the inverse of $\alpha$ so long
as $\alpha(x, x) \ne 0$ for all $x \in P$. In particular, since
$\beta(x, y)$ can be expressed in terms of $\beta(z, y)$ for $z > x$,
we may start by solving for $\beta(y, y)$ and inductively solve for
values $\beta(z, y)$ with $z < y$ until we reach $\beta(x, y)$. Note
that $\beta(x, y)$ only depends on the values of $\alpha(x', y')$ for
$x', y' \in [x, y]$.
\begin{exercise}
We have shown that $\alpha(x, x) \ne 0$ is a sufficient condition for
$\alpha$ to have an inverse. Observe that it is also necesary.
\end{exercise}
\section{Mobius inversion for posets}
We now shift our attention to two particular elements of
$I(P)$. Consider the element $\zeta \in I(P)$ given by $\zeta(x, y) =
1$ if $x \le y$ and $\zeta(x, y) = 0$ otherwise. Since $\zeta(x, x)
\ne 0$ for all $x \in P$, $\zeta$ has an inverse, which will be
denoted by $\mu$. As will be seen later, the suggestive naming of
these elements is meant to draw a connection to the zeta and Mobius
functions we have studied before in the context of Dirichlet series.
The first hint towards this connection is the Mobius inversion formula
for posets.
\begin{theorem}[Mobius inversion formula (posets)] \label{mobius inversion}
Let $P$ be a poset and $K$ a field, and let $f$ and $g$ be functions
from $P$ to $K$. If $f$ satisfies
\[ f(x) = \sum_{y \le x} g(y) \]
for all $x \in P$, then $g$ is given by
\[ g(x) = \sum_{y \le x} f(y) \mu(y, x). \]
Conversely, if the second equation holds, then so does the first.
\end{theorem}
\begin{proof}
It is possible to verify this directly by substituting the equation
for $f$ into the equation for $g$ that is to be proven. However, it
is perhaps more insightful to take a different approach.
Just as we interpreted elements of $I(P)$ as $|P| \times |P|$
matrices, for a function $f : P \to K$, we may associate to $f$ a
row vector $M(f)$ with $|P|$ columns indexed by the elements of $P$,
where the column $x$ entry is $f(x)$. It is not hard to verify,
then, that the condition
\[ f(x) = \sum_{y \le x} g(y) \]
is equivalent to the equation $M(f) = M(g)M(\zeta)$. We can then
multiply both sides on the right by $M(\mu)$, and because $\mu$ is
the inverse of $\zeta$, we obtain $M(f)M(\mu) = M(g)$. Writing this
matrix equation out entry-by-entry, we find that
\[ g(x) = \sum_{y \le x} f(y) \mu(y, x), \]
as desired. A similar argument proves the converse.
\end{proof}
The compactness of the equations in the preceding proof when we
converted to matrices should not be taken as mere algebraic
coincidence. It is sometimes fruitful to think of elements of $I(P)$
as acting on the $|P|$-dimensional vector space $K^P$ of all functions
from $P$ to $K$. We have just seen the right action of $I(P)$ on
$K^P$, giving $K^P$ the structure of a right $I(P)$-module.
In fact, it is also possible to give $K^P$ the structure of a left
$I(P)$-module by expressing the elements of $K^P$ as column vectors
instead of row vectors. This leads us to another version of the Mobius
inversion formula.
\begin{theorem}[Mobius inversion formula (dual version)] \label{dual mobius inversion}
Let $P$ be a poset and $K$ a field, and let $f$ and $g$ be functions
from $P$ to $K$. If $f$ satisfies
\[ f(x) = \sum_{y \ge x} g(y) \]
for all $x \in P$, then $g$ is given by
\[ g(x) = \sum_{y \ge x} \mu(x, y) f(y). \]
Conversely, if the second equation holds, then so does the first.
\end{theorem}
\begin{proof}
The proof follows along the same lines as the proof of Theorem
\ref{mobius inversion}. Alternatively, setting $Q = P^*$, this follows
from Mobius inversion on $Q$ upon noting that $\mu_Q(x, y) = \mu_P(y,
x)$. The proof of this last statement is left as an exercise for the
reader.
\end{proof}
\begin{example} \label{P = n}
Let us apply Mobius inversion to the case $P = {\bf n}$. In the
previous section, we outlined a procedure by which the inverse of an
element $\alpha \in I(P)$ may be computed. Applying this to $\alpha =
\zeta$, we find that $\mu(x, x) = 1$, $\mu(x, x + 1) = -1$, and
$\mu(x, y) = 0$ for $y > x + 1$.
Plugging this into the Mobius inversion formula yields
\[ f(k) = g(1) + \ldots + g(k) \iff g(k) = \begin{cases} f(k) - f(k - 1) & \text{if $k \ge 2$} \\ f(1) & \text{if $k = 1$} \end{cases}, \]
which expresses a familiar relationship between a series and its
partial sums.
\end{example}
\begin{exercise}
In the above example, compute $\mu$ instead by computing the matrix
inverse of $M(\zeta)$. (It may be helpful to think of $M(\zeta)$ as a
change of basis for the vector space of polynomials of degree less
than $n$.)
\end{exercise}
It was fairly easy to compute $\mu$ in the last example, but for more
complicated posets, it may not be so simple. However, in the case of
products of posets, the following lemma can make the computation more
tractable.
\begin{lemma} \label{product formula}
Let $P$ and $Q$ be posets, and consider any elements $x, x' \in P$ and
$y, y' \in Q$, where $x \le x'$ and $y \le y'$. Then,
\[ \mu_{P \times Q} \left( (x, y), (x', y') \right) = \mu_P(x, x') \mu_Q(y, y'), \]
where the subscript on $\mu$ indicates which poset is being used to
define the Mobius function.
\end{lemma}
\begin{proof}
Let $\mu' \left( (x, y), (x', y') \right) = \mu_P(x, x') \mu_Q(y,
y')$. We may compute
\[ \sum_{(x, y) \le (u, v) \le (x', y')} \mu' \left( (x, y), (u, v) \right) \]
\[ = \sum_{(x, y) \le (u, v) \le (x', y')} \mu_P(x, u) \mu_Q(y, v) \]
\[ = \left( \sum_{x \le u \le x'} \mu_P(x, u) \right) \left( \sum_{y \le v \le y'} \mu_Q(y, v) \right) \]
\[ = \delta_{xx'} \delta_{yy'} = \delta_{(x, y)(x', y')}. \]
This equation says exactly that $\mu'$ is the inverse of $\zeta_{P
\times Q}$, so $\mu_{P \times Q} = \mu'$.
\end{proof}
It may be somewhat surprising that the above calculation worked out so
conveniently. Whenever calculations work out nicely in mathematics,
there is some hope that a more abstract underlying theory is at
work. Such is the case here, and we shall take a digression to develop
briefly the abstract viewpoint. (The uninterested reader may skip
ahead.)
\section{A digression on tensor products}
Before we define the tensor product of two vector spaces, it will be
useful to review the concept of a dual vector space. If $V$ is a
$K$-vector space, then we define $V^*$ to be the space of all linear
maps from $V$ to $K$, and we call $V^*$ the \emph{dual} of $V$. In the
case that $V$ is finite-dimensional, it is not hard to check that
$V^*$ has the same dimension as $V$.
Since all vector spaces of the same finite dimension are isomorphic,
defining $V^*$ in this way may not seem like a particularly useful
notion. However, an isomorphism between two arbitrary vector spaces of
the same finite dimension is not necessarily \emph{natural}: it
requires picking a basis for each vector space, and a different choice
of bases will result in a different isomorphism. If the definition of
tensor product depended on a choice of basis, then it would be nothing
more than a computational tool. Instead, we shall see that the tensor
product can be defined in an intrinsic way; we may then specialize to
particular bases to reap the insight gained from the abstract
viewpoint.
Suppose now that $W$ is another vector space, and $T$ is a linear map
from $V$ to $W$. Then, we can associate to $T$ a linear map $T^* : W^*
\to V^*$ defined by $T^*(f)(v) = f(T(v))$ for $f \in W^*$ and $v \in
V$. The reader should think of $T$ and $T^*$ as two sides of the same
coin. If $S$ is a map from another vector space $U$ to $V$, then it is
not hard to check that $(ST)^* = T^*S^*$.
Given vector spaces $V$ and $W$, define the \emph{tensor product} $V
\otimes W$ to be the space of all bilinear forms on $V^* \times
W^*$. If $S : V \to V$ and $T : W \to W$ are linear maps, we can
define a linear map $S \otimes T : V \otimes W \to V \otimes W$ as
follows: for $B \in V \otimes W$, we define
\[ ((S \otimes T) B))(f, g) = B(S^*f, T^*g) \]
for all $f \in V^*$ and $g \in W^*$. The use of dual spaces here seems
a bit onerous, but there is a good reason for it. Suppose that we have
two more maps $S' : V \to V$ and $T' : W \to W$. It is not hard to
check from the definitions that
\[ (S' \otimes T') \circ (S \otimes T) = (S'S \otimes T'T). \]
If we had instead defined $V \otimes W$ as the space of bilinear forms
on $V \times W$ and proceeded in an analogous way, this composition
law would be reversed.
Now that we have established some basic properties of the tensor
product, it is useful to see what it looks like in coordinates. Let
$v_1, \ldots , v_n$ be a basis of $V$, and let $w_1, \ldots, w_m$ be a
basis of $W$. Let $v^*_1, \ldots , v^*_n$ be the corresponding basis
for $V^*$, where $v^*_i(v_j) = 1$ if $j = i$ and $v^*_i(v_j) = 0$
otherwise. Define $w^*_1, \ldots , w^*_m$ in an analogous manner, and
let $B_{ij} \in V \otimes W$ be the bilinear form given by
$B_{ij}(v^*_{i'}, w^*_{j'}) = 1$ if $(i, j) = (i', j')$ and
$B_{ij}(v^*_{i'}, w^*_{j'}) = 0$ otherwise.
The $B_{ij}$ form a basis of $V \otimes W$, and we will analyze $S
\otimes T$ in the $B_{ij}$ coordinates. Note that
\[ ((S \otimes T) B_{ij})(v^*_{i'}, w^*_{j'}) = B_{ij}(S^*(v^*_{i'}), T^*(w^*_{j'})) \]
\[ = \left( S^*(v^*_{i'})(v_i) \right) \left( T^*(w^*_{j'})(w_j) \right) \]
\[ = v^*_{i'}(Sv_i) w^*_{j'}(Tw_j) = S_{i'i}T_{j'j}. \]
The first expression is the $B_{i'j'}$ component of $(S \otimes T)
B_{ij}$, so we conclude that the matrix entries of $S \otimes T$ are
products of matrix entries of $S$ and $T$; in particular, $(S \otimes
T)_{(i', j')(i, j)} = S_{i'i} T_{j'j}$.
We can apply the theory developed thus far to the case of the $\zeta$
and $\mu$ functions on product posets. For posets $P$ and $Q$, it is
trivial to check that
\[ \zeta_{P \times Q} \left( (x, y), (x', y') \right) = \zeta_P(x, x') \zeta_Q(y, y'). \]
Thus, $M(\zeta_{P \times Q}) = M(\zeta_P) \otimes M(\zeta_Q)$. From
the composition properties we have shown for tensor products of maps,
we can invert this equation to obtain
\[ M(\zeta_{P \times Q})^{-1} = M(\zeta_P)^{-1} \otimes M(\zeta_Q)^{-1} \]
\[ M(\mu_{P \times Q}) = M(\mu_P) \otimes M(\mu_Q). \]
In coordinates, this is
\[ \mu_{P \times Q} \left( (x, y), (x', y') \right) = \mu_P(x, x') \mu_Q(y, y'), \]
which gives us another proof of Lemma \ref{product formula}. This was
the purported purpose of our digression. Of course, it may seem rather
anticlimactic for all the theory developed here to culminate in a
result that can be proven by calculation within a few lines. However,
the theory of tensor products\footnote{Note: We have only developed
the theory of tensor products of vector spaces. More generally, one
can define tensor products of modules.} is not limited to the
context of posets and incidence algebras. Tensor products appear in
diverse areas of mathematics and once learned, are often a useful tool
for understanding results that may otherwise seem unmotivated.
\begin{exercise}
Let $V$ and $W$ be vector spaces. Suppose that we specify bases on $V$
and $W$; as we saw above, these give rise naturally to corresponding
bases for $V^*$ and $W^*$. With these bases, we can write $T : V \to
W$ and $T^* : W^* \to V^*$ as matrices. Describe the relationship
between these two matrices.
\end{exercise}
\begin{exercise}
Let $T : V \to W$ be a linear map, and let $\pi : W \to W/T(V)$ be the
projection.
\begin{enumerate}
\item Show that $\pi \circ T$ is the zero map, and conclude that $(\pi
\circ T)^*$ is also the zero map.
\item Show that for any surjective linear map $S$, $S^*$ is
injective. In particular, $\pi^*$ is injective.
\item Based on the previous two parts, compute a lower bound on the
dimension of the image of $T^*$ in terms of the dimensions of $V$,
$W$, and $W/T(V)$.
\item Using the previous exercise and a symmetry argument, deduce that
the row and column ranks of a matrix are equal.
\end{enumerate}
\end{exercise}
\begin{exercise}
Suppose that $S : V \to V'$ and $T : W \to W'$ are linear maps between
vector spaces. Come up with a definition of $S \otimes T$ and state a
composition law for it.
\end{exercise}
\begin{exercise}
We derived Lemma \ref{product formula} using the tensor product $V
\otimes W$, which we defined as the space of bilinear forms on $V^*
\times W^*$. Come up with an alternate proof using instead the space
of bilinear forms on $V \times W$. (In fact, this proof is perhaps
simpler than the one we have given, but it does not directly introduce
the tensor product.)
\end{exercise}
\section{Returning to posets}
Let us now return to our main topic, the Mobius function on
posets. Lemma \ref{product formula} gives us a convenient way to
compute the Mobius functions for some more complicated posets, as in
the next example.
\begin{example}
Let $P = B_n = {\bf 2} \times \cdots \times {\bf 2}$. The poset $P$
can be naturally thought of as subsets of $[n]$ under inclusion; a
subset $S \subset [n]$ represents the element $(b_1, \ldots , b_n) \in
P$, where $b_i = 2$ if $i \in S$ and $b_i = 1$ if $i \not \in S$.
If $T \subset [n]$ represents $(c_1, \ldots , c_n) \in P$, and $S
\subset T$, then we have by Lemma \ref{product formula}
\[ \mu_P(S, T) = \mu_P \left( (b_1, \ldots , b_n) , (c_1, \ldots , c_n) \right) \]
\[ = \prod_{i = 1}^n \mu_{\bf 2}(b_i, c_i) \]
\[ = (-1)^{|T| - |S|}, \]
where we have used the formula for $\mu_{\bf 2}$ from Example \ref{P =
n}. Applying Mobius inversion, we find that
\[ f(T) = \sum_{S \subset T} g(S) \iff g(T) = \sum_{S \subset T} (-1)^{|T| - |S|} f(S). \]
This formula reminds us of the principle of inclusion-exclusion, and
in fact, it is a generalization. To derive the principle of
inclusion-exclusion as a special case, however, it is more convenient
to work with Theorem \ref{dual mobius inversion}, which states that
\[ f(T) = \sum_{S \supset T} g(S) \iff g(T) = \sum_{S \supset T} (-1)^{|S| - |T|} f(S). \]
To enter the setting of inclusion-exclusion, suppose we have sets
$E_1, \ldots , E_n$, and we wish to compute the cardinality of $X =
E_1 \cup \cdots \cup E_n$. Let $g(S)$ be the number of elements in $X$
which are contained in $E_s$ if and only if $s \in S$. If $f(T) =
\sum_{S \supset T} g(S)$, then $f(T)$ counts the number of elements in
$\bigcup_{t \in T} E_t$. Rearranging the inversion formula gives
\[ f(\emptyset) = g(\emptyset) + \sum_{S \ne \emptyset} (-1)^{|S| + 1} f(S). \]
Noting that $f(\emptyset) = |X|$ and $g(\emptyset) = 0$, this gives us
the same formula for $|X|$ as the principle of inclusion-exclusion.
\end{example}
\begin{exercise}
Derive the principle of inclusion-exclusion directly from the first
Mobius inversion formula in the above example. Can you derive Theorem
\ref{dual mobius inversion} directly from Theorem \ref{mobius
inversion}?
\end{exercise}
\begin{example}
Let $P = D_n$ be the poset of divisors of an integer $n$, partially
ordered by divisibility. If $n$ has prime factorization $p_1^{n_1}
\cdots p_k^{n_k}$, then $P \cong ({\bf n_1 + 1}) \times \cdots \times
({\bf n_k + 1})$.
Using Lemma \ref{product formula} and Example \ref{P = n}, we find
that $\mu_P(1, n) = \mu(n)$, where the second $\mu$ is the familiar
number theoretic Mobius function defined on natural numbers. For $d_1,
d_2 \in P$ with $d_1 | d_2$, we find that $[d_1, d_2] \cong
D_{d_2/d_1}$. Thus, $\mu_P(d_1, d_2) = \mu_{D_{d_2/d_1}}(1, d_2/d_1) =
\mu(d_2/d_1)$.
This shows the connection between the Mobius function for posets and
the number theoretic Mobius function; in some sense, the number
theoretic $\zeta$ and Mobius functions are the infinite versions of
the functions of the same name we have defined on finite posets. (The
analogy is slightly off, since the Riemann $\zeta$ function refers to
a Dirichlet series, while the Mobius function refers to the
coefficients on the Dirichlet series inverse of $\zeta$.)
\end{example}
These examples conclude our basic development of the Mobius function
for posets. We end with a lemma that will be used in the next lecture
and also revisits the chain counting results from the previous
lecture.
\begin{lemma}
Let $P$ be a finite rank $n$ poset, and let $\widehat{P} = {\bf 1}
\oplus P \oplus {\bf 1}$. Denote the minimal and maximal elements of
$\widehat{P}$ by $\widehat{0}$ and $\widehat{1}$, respectively. If
$c_i$ is the number of length $i$ chains from $\widehat{0}$ to
$\widehat{1}$, then
\[ \mu_{\widehat{P}} (\widehat{0}, \widehat{1}) = \sum_{i = 0}^{n + 1} (-1)^i c_i, \]
where we define $c_0 = 1$.
\end{lemma}
\begin{proof}
Recall from the previous lecture that $(\zeta_{\widehat{P}} - 1)^k(x,
y)$ counts the number of length $k$ chains from $x$ to $y$. In
particular, $(\zeta_{\widehat{P}} - 1)^k(\widehat{0}, \widehat{1}) =
c_k$.
This provides the connection between $\mu_{\widehat{P}}$ and the
$c_i$. We can expand $\mu_{\widehat{P}}$ as a power series as follows.
\[ \mu_{\widehat{P}} = \left( 1 + (\zeta_{\widehat{P}} - 1) \right)^{-1} \]
\[ = \sum_{i = 0}^\infty (-1)^i (\zeta_{\widehat{P}} - 1)^i. \]
The sum is actually finite, since $\zeta_{\widehat{P}} - 1$ is
nilpotent. Evaluating both sides at $[\widehat{0}, \widehat{1}]$ and
applying our initial observation yields the lemma.
\end{proof}
\end{document}