Background
I'm currently working with incidence algebras on posets. An incidence algebra was defined as $\{\text{Interval functions on } P\}$ but this notion is the same as defining the algebra as
$$\mathscr{I}(P)=\{f:P^2\to\mathbb C:\ (x>_Py)\Rightarrow (f(x,y)=0)\}$$
where $>_P$ is the negation of the order $(P,\leq_P)$. I know that $\mathscr{I}(P)$ is a $\mathbb C$-algebra, but I haven't proved it yet.
The multiplication operation on this set is defined by
$$(fg)(x,y)=\sum_{x\leq z\leq y}f(x,z)g(z,y)$$ and this operation is not commutative in general. There's an identity element $\delta(x,y)$ a function which behaves just like the Kronecker-Delta function $1$ when $x=y$ and $0$ otherwise.
The Statement
I am told to prove that
$fg=\delta\iff gf=\delta$.
That is, if $g$ is right-inverse, then it's a left-inverse for $f$.
My Approach
Initially this problem comes after we have been asked to prove that
$f$ is invertible $\iff\ \forall x\in P(f(x,x)\neq 0)$
The proof of this fact can be found in pages 12-13 of the book Incidence Algebras by Spiegel and O'Donnel, however the proof doesn't rely on the fact $f$ is invertible on both sides. Instead it proves the equivalence between the latter fact and $f$ being right-invertible.
It follows using the same idea that $f$ is left-invertible when $f(x,x)\neq 0$ holds. So I remember that existence of a left-inverse and a right-inverse implies that they must be the same when the operation is associative. So a good proof would be
$$fg=\delta \iff f(x,x)\neq 0\iff hf=\delta$$
and thus $g=h$ as lateral inverses must coincide and then $fg=gf$ where $g=f^{-1}$, however I'm at an impasse.
My Problem
Even if I have stated that the incidence algebra $\mathscr{I}(P)$ is indeed an algebra, I haven't proven it. My proof isn't accepted because I claimed the operation was associative but haven't shown it.
On the other hand, I can take the right inverse $g$ obtained from the proof in Spiegel and O'Donnel's book and compute the product $gf$ and see it equals the $\delta$ function. However this calculation has been difficult and I can't see how to come around it.
In the book, the function $g(x,y)$ is defined inductively on the size on the interval $[x,y]$ as follows:
$$g(x,y)=\frac{-1}{f(x,x)}\sum_{x<z\leq y}f(x,z)g(z,y).$$
Even when taking the product $fg$, I can't see how that is the $\delta$ function. On the other side, I computed $gf$ as follows:
\begin{align*} (gf)(x,y)&=\sum_{x\leq z\leq y}g(x,z)f(z,y)\\ &=\sum_{x\leq z\leq y}\left(\frac{-1}{f(x,x)}\sum_{x< t\leq z}f(x,t)g(t,z)\right)f(z,y)\\ &=\sum_{x\leq z\leq y}\left(\frac{-1}{f(x,x)}\left(\sum_{x< t< z}f(x,t)g(t,z)+f(x,z)g(z,z)\right)\right)f(z,y) \end{align*}
where in the last step I separated the last term of the sum. I realize that this process might be just an abstract multiplication of symbols which represent complex numbers and in the end $g(x,z)$ where $[x,z]$ is an interval of smaller size than $[x,y]$ represents a complex number, not $\frac{1}{f(x,z)}$. When $x=y$ most of the sums collapse and we are left with a fraction which reduces to $1$ so in that sense $gf$ behaves like $\delta$.
My Question
In essence:
- How do I finish the proof for when $x<y$? Are my calculation fruitless?
- Am I better off just proving that $(fg)h=f(gh)$, for all elements in $\mathscr{I}(P)$. and then finishing the proof using uniqueness of lateral-inverses? Won't the calculation for associativity be much worse?
Any type of lead, suggestion or comment will be kindly received. Thank you in advance!
Let $f,g \in \mathscr{I}(P)$ such that for all $x,y \in P$ $$fg(x,y) =\sum_{x\leq z\leq y}f(x,z)g(z,y) =\delta(x,y).$$ We will prove that $gf = \delta$.
Note that if either $x,y$ are incomparable or $x>y$, then the sum is empty and $fg(x,y)=0$. If $x=y$ then $$1=fg(x,y)= f(x,x)g(y,y)=f(y,y)g(x,x)=gf(x,y).$$ The only non-trivial case is when $x<y$.
I will assume that the interval $[x,y]= \{ z \in P | x\leq z \leq y\}$ is finite (I don't have a proof for the general case, even assuming convergence), hence let $$[x,y]= \{ x=z_1,z_2,z_3,\dots,z_n=y\}.$$
Define $n \times n$ matrices $F,G$ given by $$[F]_{ij}=f(z_i,z_j) $$ $$[G]_{ij}=g(z_i,z_j).$$
Considering the matrix product, note that $$[FG]_{ij}= \sum_{k=1}^n f(z_i,z_k)g(z_k,z_j)=\sum_{w \in [z_i,z_j]}f(z_i,w)g(w,z_j)= (fg)(z_i,z_j).$$ The second to last equality comes from the fact that $z_k \not \in [z_i,z_j] \Rightarrow f(z_i,z_k)g(z_k,z_j)=0.$
Finally we obtain $$fg=\delta \iff FG = I \iff GF=I \iff gf=\delta.$$