Lindström–Gessel–Viennot lemma

Let G be a locally finite directed acyclic graph. This means that each vertex has finite degree, and that G contains no directed cycles. Consider base vertices ${\displaystyle A=\{a_{1},\ldots ,a_{n}\}}$ and destination vertices ${\displaystyle B=\{b_{1},\ldots ,b_{n}\}}$, and also assign a weight ${\displaystyle \omega _{e}}$ to each directed edge e. These edge weights are assumed to belong to some commutative ring. For each directed path P between two vertices, let ${\displaystyle \omega (P)}$ be the product of the weights of the edges of the path. For any two vertices a and b, write e(a,b) for the sum ${\displaystyle e(a,b)=\sum _{P:a\to b}\omega (P)}$ over all paths from a to b. This is well-defined if between any two points there are only finitely many paths; but even in the general case, this can be well-defined under some circumstances (such as all edge weights being pairwise distinct formal indeterminates, and ${\displaystyle e(a,b)}$ being regarded as a formal power series). If one assigns the weight 1 to each edge, then e(a,b) counts the number of paths from a to b.

With this setup, write

${\displaystyle M={\begin{pmatrix}e(a_{1},b_{1})&e(a_{1},b_{2})&\cdots &e(a_{1},b_{n})\\e(a_{2},b_{1})&e(a_{2},b_{2})&\cdots &e(a_{2},b_{n})\\\vdots &\vdots &\ddots &\vdots \\e(a_{n},b_{1})&e(a_{n},b_{2})&\cdots &e(a_{n},b_{n})\end{pmatrix}}}$.

An n-tuple of non-intersecting paths from A to B means an n-tuple (P₁, ..., P_n) of paths in G with the following properties:

• There exists a permutation ${\displaystyle \sigma }$ of ${\displaystyle \left\{1,2,...,n\right\}}$ such that, for every i, the path P_i is a path from ${\displaystyle a_{i}}$ to ${\displaystyle b_{\sigma (i)}}$.
• Whenever ${\displaystyle i\neq j}$, the paths P_i and P_j have no two vertices in common (not even endpoints).

Given such an n-tuple (P₁, ..., P_n), we denote by ${\displaystyle \sigma (P)}$ the permutation of ${\displaystyle \sigma }$ from the first condition.

The Lindström–Gessel–Viennot lemma then states that the determinant of M is the signed sum over all n-tuples P = (P₁, ..., P_n) of non-intersecting paths from A to B:

${\displaystyle \det(M)=\sum _{(P_{1},\ldots ,P_{n})\colon A\to B}\mathrm {sign} (\sigma (P))\prod _{i=1}^{n}\omega (P_{i}).}$

That is, the determinant of M counts the weights of all n-tuples of non-intersecting paths starting at A and ending at B, each affected with the sign of the corresponding permutation of ${\displaystyle (1,2,\ldots ,n)}$, given by ${\displaystyle P_{i}}$ taking ${\displaystyle a_{i}}$ to ${\displaystyle b_{\sigma (i)}}$.

In particular, if the only permutation possible is the identity (i.e., every n-tuple of non-intersecting paths from A to B takes a_i to b_i for each i) and we take the weights to be 1, then det(M) is exactly the number of non-intersecting n-tuples of paths starting at A and ending at B.

To prove the Lindström–Gessel–Viennot lemma, we first introduce some notation.

An n-path from an n-tuple ${\displaystyle (a_{1},a_{2},\ldots ,a_{n})}$ of vertices of G to an n-tuple ${\displaystyle (b_{1},b_{2},\ldots ,b_{n})}$ of vertices of G will mean an n-tuple ${\displaystyle (P_{1},P_{2},\ldots ,P_{n})}$ of paths in G, with each ${\displaystyle P_{i}}$ leading from ${\displaystyle a_{i}}$ to ${\displaystyle b_{i}}$. This n-path will be called non-intersecting just in case the paths P_i and P_j have no two vertices in common (including endpoints) whenever ${\displaystyle i\neq j}$. Otherwise, it will be called entangled.

Given an n-path ${\displaystyle P=(P_{1},P_{2},\ldots ,P_{n})}$, the weight ${\displaystyle \omega (P)}$ of this n-path is defined as the product ${\displaystyle \omega (P_{1})\omega (P_{2})\cdots \omega (P_{n})}$.

A twisted n-path from an n-tuple ${\displaystyle (a_{1},a_{2},\ldots ,a_{n})}$ of vertices of G to an n-tuple ${\displaystyle (b_{1},b_{2},\ldots ,b_{n})}$ of vertices of G will mean an n-path from ${\displaystyle (a_{1},a_{2},\ldots ,a_{n})}$ to ${\displaystyle \left(b_{\sigma (1)},b_{\sigma (2)},\ldots ,b_{\sigma (n)}\right)}$ for some permutation ${\displaystyle \sigma }$ in the symmetric group ${\displaystyle S_{n}}$. This permutation ${\displaystyle \sigma }$ will be called the twist of this twisted n-path, and denoted by ${\displaystyle \sigma (P)}$ (where P is the n-path). This, of course, generalises the notation ${\displaystyle \sigma (P)}$ introduced before.

Recalling the definition of M, we can expand det M as a signed sum of permutations; thus we obtain

${\displaystyle {\begin{array}{rcl}\det M&=&\sum _{\sigma \in S_{n}}\mathrm {sign} (\sigma )\prod _{i=1}^{n}e(a_{i},b_{\sigma (i)})\\&=&\sum _{\sigma \in S_{n}}\mathrm {sign} (\sigma )\prod _{i=1}^{n}\sum _{P_{i}:a_{i}\to b_{\sigma (i)}}\omega (P_{i})\\&=&\sum _{\sigma \in S_{n}}\mathrm {sign} (\sigma )\sum \{\omega (P):P~{\text{an}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{\sigma (1)},b_{\sigma (2)},\ldots ,b_{\sigma (n)}\right)\}\\&=&\sum \{\mathrm {sign} (\sigma (P))\omega (P):P~{\text{a twisted}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{1},b_{2},...,b_{n}\right)\}\\&=&\sum \{\mathrm {sign} (\sigma (P))\omega (P):P~{\text{a non-intersecting twisted}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{1},b_{2},...,b_{n}\right)\}\\&&+\sum \{\mathrm {sign} (\sigma (P))\omega (P):P~{\text{an entangled twisted}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{1},b_{2},...,b_{n}\right)\}\\&=&\sum _{(P_{1},\ldots ,P_{n})\colon A\to B}\mathrm {sign} (\sigma (P))\omega (P)\\&&+\underbrace {\sum \{\mathrm {sign} (\sigma (P))\omega (P):P~{\text{an entangled twisted}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{1},b_{2},...,b_{n}\right)\}} _{=0?}\\\end{array}}}$

It remains to show that the sum of ${\displaystyle \mathrm {sign} (\sigma (P))\omega (P)}$ over all entangled twisted n-paths vanishes. Let ${\displaystyle {\mathcal {E}}}$ denote the set of entangled twisted n-paths. To establish this, we shall construct an involution${\displaystyle f:{\mathcal {E}}\longrightarrow {\mathcal {E}}}$ with the properties ${\displaystyle \omega (f(P))=\omega (P)}$ and ${\displaystyle \mathrm {sign} (\sigma (f(P)))=-\mathrm {sign} (\sigma (P))}$ for all ${\displaystyle P\in {\mathcal {E}}}$. Given such an involution, the rest-term

${\displaystyle \sum \{\mathrm {sign} (\sigma (P))\omega (P):P~{\text{an entangled twisted}}~n{\text{-path from}}~\left(a_{1},a_{2},\ldots ,a_{n}\right)~{\text{to}}~\left(b_{1},b_{2},...,b_{n}\right)\}=\sum _{P\in {\mathcal {E}}}\mathrm {sign} (\sigma (P))\omega (P)}$

in the above sum reduces to 0, since its addends cancel each other out (namely, the addend corresponding to each ${\displaystyle P\in {\mathcal {E}}}$ cancels the addend corresponding to ${\displaystyle f(P)}$).

Construction of the involution: The idea behind the definition of the involution ${\displaystyle f}$ is to take choose two intersecting paths within an entangled path, and switch their tails after their point of intersection. There are in general several pairs of intersecting paths, which can also intersect several times; hence, a careful choice needs to be made. Let ${\displaystyle P=\left(P_{1},P_{2},...,P_{n}\right)}$ be any entangled twisted n-path. Then ${\displaystyle f(P)}$ is defined as follows. We call a vertex crowded if it belongs to at least two of the paths ${\displaystyle P_{1},P_{2},...,P_{n}}$. The fact that the graph is acyclic implies that this is equivalent to "appearing at least twice in all the paths".[1] Since P is entangled, there is at least one crowded vertex. We pick the smallest ${\displaystyle i\in \{1,2,\ldots ,n\}}$ such that ${\displaystyle P_{i}}$ contains a crowded vertex. Then, we pick the first crowded vertex v on ${\displaystyle P_{i}}$ ("first" in sense of "encountered first when travelling along ${\displaystyle P_{i}}$"), and we pick the largest j such that v belongs to ${\displaystyle P_{j}}$. The crowdedness of v implies j > i. Write the two paths ${\displaystyle P_{i}}$ and ${\displaystyle P_{j}}$ as

${\displaystyle {\begin{array}{rcl}P_{i}&\equiv &a_{i}=u_{0}\to u_{1}\to u_{2}\ldots u_{\alpha -1}\to \overbrace {\mathbf {u} _{\alpha }\to u_{\alpha +1}\ldots \to u_{r}=b_{\sigma (i)}} ^{\mathrm {tail} ~i}\\P_{j}&\equiv &a_{j}=v_{0}\to v_{1}\to v_{2}\ldots v_{\beta -1}\to \underbrace {\mathbf {v} _{\beta }\to v_{\beta +1}\ldots \to v_{s}=b_{\sigma (j)}} _{\mathrm {tail} ~j}\\\end{array}}}$

where ${\displaystyle \sigma =\sigma (P)}$, and where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are chosen such that v is the ${\displaystyle \alpha }$-th vertex along ${\displaystyle P_{i}}$ and the ${\displaystyle \beta }$-th vertex along ${\displaystyle P_{j}}$ (that is, ${\displaystyle v=u_{\alpha }=v_{\beta }}$). We set ${\displaystyle \alpha _{P}=\alpha }$ and ${\displaystyle \beta _{P}=\beta }$ and ${\displaystyle i_{P}=i}$ and ${\displaystyle j_{P}=j}$. Now define the twisted n-path ${\displaystyle f(P)}$ to coincide with ${\displaystyle P}$ except for components ${\displaystyle i}$ and ${\displaystyle j}$, which are replaced by

${\displaystyle {\begin{array}{rcl}P'_{i}&\equiv &a_{i}=u_{0}\to u_{1}\to u_{2}\ldots u_{\alpha -1}\to \overbrace {v_{\beta _{P}}\to v_{\beta _{P}+1}\ldots \to v_{s}=b_{\sigma (j)}} ^{\mathrm {tail} ~j}\\P'_{j}&\equiv &a_{j}=v_{0}\to v_{1}\to v_{2}\ldots v_{\beta -1}\to \underbrace {u_{\alpha _{P}}\to u_{\alpha _{P}+1}\ldots \to u_{r}=b_{\sigma (i)}} _{\mathrm {tail} ~i}\\\end{array}}}$

It is immediately clear that ${\displaystyle f(P)}$ is an entangled twisted n-path. Going through the steps of the construction, it is easy to see that ${\displaystyle i_{f(P)}=i_{P}}$, ${\displaystyle j_{f(P)}=j_{P}}$ and furthermore that ${\displaystyle \alpha _{f(P)}=\alpha _{P}}$ and ${\displaystyle \beta _{f(P)}=\beta _{P}}$, so that applying ${\displaystyle f}$ again to ${\displaystyle f(P)}$ involves swapping back the tails of ${\displaystyle f(P)_{i},f(P)_{j}}$ and leaving the other components intact. Hence ${\displaystyle f(f(P))=P}$. Thus ${\displaystyle f}$ is an involution. It remains to demonstrate the desired antisymmetry properties:

From the construction one can see that ${\displaystyle \sigma (f(P))}$ coincides with ${\displaystyle \sigma =\sigma (P)}$ except that it swaps ${\displaystyle \sigma (i)}$ and ${\displaystyle \sigma (j)}$, thus yielding ${\displaystyle \mathrm {sign} (\sigma (f(P)))=-\mathrm {sign} (\sigma (P))}$. To show that ${\displaystyle \omega (f(P))=\omega (P)}$ we first compute, appealing to the tail-swap

${\displaystyle {\begin{array}{rcl}\omega (P'_{i})\omega (P'_{j})&=&\left(\prod _{t=0}^{\alpha -1}\omega (u_{t},u_{t+1})\cdot \prod _{t=\beta }^{s-1}\omega (v_{t},v_{t+1})\right)\cdot \left(\prod _{t=0}^{\beta -1}\omega (v_{t},v_{t+1})\cdot \prod _{t=\alpha }^{r-1}\omega (u_{t},u_{t+1})\right)\\&=&\prod _{t=0}^{r-1}\omega (u_{t},u_{t+1})\cdot \prod _{t=0}^{s-1}\omega (v_{t},v_{t+1})\\&=&\omega (P_{i})\omega (P_{j}).\\\end{array}}}$

Hence ${\displaystyle \omega (f(P))=\prod _{k=1}^{n}\omega (f(P)_{k})=\prod _{k=1,~k\neq i,j}^{n}\omega (P_{k})\cdot \omega (P'_{i})\omega (P'_{j})=\prod _{k=1,~k\neq i,j}^{n}\omega (P_{k})\cdot \omega (P_{i})\omega (P_{j})=\prod _{k=1}^{n}\omega (P_{k})=\omega (P)}$.

Thus we have found an involution with the desired properties and completed the proof of the Lindström-Gessel-Viennot lemma.

Remark. Arguments similar to the one above appear in several sources, with variations regarding the choice of which tails to switch. A version with j smallest (unequal to i) rather than largest appears in the Gessel-Viennot 1989 reference (proof of Theorem 1).

• Matrix tree theorem

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• Sagan, Bruce E. (2001), The symmetric group, Springer
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• Talaska, Kelli (2012), Determinants of weighted path matrices, arXiv:1202.3128v1
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• Fulmek, Markus (2010), Viewing determinants as nonintersecting lattice paths yields classical determinantal identities bijectively, arXiv:1010.3860v1