I would like to verify that every Grassmann graph is distance-transitive.

Suppose that we are given arbitrary natural number \(n\), a natural number \(k \leq n\), and a prime power \(q\). Let \(S = \{ W : W \leq V \text{ and } \operatorname{dim}(W) = k \}\) denote the set of \(k\)-dimensional subspaces of a vector space \(V\) over the finite field \(\mathbb{F}_{q}\) of order \(q\). Consider the simple graph \(G=(S,E)\) formed with the set \(S\) as vertices and the set \(E=\{\{W,Z\} : W,Z \in S \operatorname{and} \operatorname{dim}(W \cap Z)= k - 1\}\) as edges. Then:

  1. Suppose that \(L : V \to V\) is a linear isomorphism.

    Let \(F : S \to S\) be a function on the vertices of \(G\) where \(F(U) := \operatorname{image}(L\vert_U)\) for all \(U \in S\).

    Now, select an arbitrary pair of vertices \(X,Y \in S\) such that \(X\) is adjacent to \(Y\) on \(G\).

    \[\implies \operatorname{dim}(F(X) \cap F(Y))\] \[= \operatorname{dim}(\operatorname{image}(L\vert_X) \cap \operatorname{image}(L \vert_Y))\] \[= \operatorname{dim}(\{ v : v = L(x) = L(y) \text { for some } x \in X \text{ and } y \in Y\})\] \[= \operatorname{dim}(\{ v : L^{-1}(v) = x = y \text { for some } x \in X \text{ and } y \in Y\})\] \[= \operatorname{dim}(\{ v : L^{-1}(v) = z \text { for some } z \in X \cap Y \})\] \[= \operatorname{dim}(\{ v : v = L(z) \text { for some } z \in X \cap Y \})\] \[= \operatorname{dim}(\operatorname{image}(X \cap Y))\] \[= k - 1\] \[\implies F(X) \text{ is adjacent to } F(Y)\] \[\implies F \in \operatorname{Aut}(G) \text{ is an automorphism on } G\]

    [ since \(X\) and \(Y\) are arbitrary, defining \(F^{-1} : S \to S\) as \(F^{-1}(U) := \operatorname{image}(L^{-1}\vert_U)\) for all \(U \in S\)].

  2. Suppose that \(W_1, W_2, X_1, X_2 \in S\) such that \(d(W_1,W_2)=d(X_1,X_2)\). Then:

    \(W_1 = (W_1 \cap W_2) \oplus Y_{W_2}\) and \(W_2 = (W_1 \cap W_2) \oplus Y_{W_1}\)

    [ for some \(Y_{W_2} \leq W_1\) where \(Y_{W_2} \cap W_2 = \{0\}\), and some \(Y_{W_1} \leq W_2\) where \(Y_{W_1} \cap W_1 = \{0\}\) ]

    \[\implies V = (W_1 + W_2) \oplus Z_{W_1 + W_2} = (W_1 \cap W_2) \oplus Y_{W_1} \oplus Y_{W_2} \oplus Z_{W_1 + W_2}\]

    [ for some \(Z_{W_1 + W _2} \leq V\) where \(Z_{W_1 + W_2} \cap (W_1 + W_2) = \{0\}\) ]

    Similarly, it must be the case that:

    \(X_1 = (X_1 \cap X_2) \oplus Y_{X_2}\) and \(X_2 = (X_1 \cap X_2) \oplus Y_{X_1}\)

    [ for some \(Y_{X_2} \leq X_1\) where \(Y_{X_2} \cap X_2 = \{0\}\), and some \(Y_{X_1} \leq X_2\) where \(Y_{X_1} \cap X_1 = \{0\}\) ]

    \[\implies V = (X_1 + X_2) \oplus Z_{X_1 + X_2} = (X_1 \cap X_2) \oplus Y_{X_1} \oplus Y_{X_2} \oplus Z_{X_1 + X_2}\]

    [ for some \(Z_{X_1 + X _2} \leq V\) where \(Z_{X_1 + X_2} \cap (X_1 + X_2) = \{0\}\) ]

    Now, note that \(\operatorname{dim}(W_1 \cap W_2) = \operatorname{dim}(X_1 \cap X_2)\), \(\operatorname{dim}(Y_{W_1}) = \operatorname{dim}(Y_{W_2}) = \operatorname{dim}(Y_{X_1}) = \operatorname{dim}(Y_{X_1})\), and \(\operatorname{dim}(Z_{W_1 + W_2}) = \operatorname{dim}(Z_{X_1 + X_2})\) - since \(d(W_1,W_2)=d(X_1,X_2)\) on \(G\).

    This means here that it is possible to construct a linear isomorphism \(L : V \to V\) such that \(\operatorname{image}(L\vert_{W_1 \cap W_2}) = X_1 \cap X_2\), \(\operatorname{image}(L\vert_{Y_{W_1}}) = Y_{X_1}\), \(\operatorname{image}(L\vert_{Y_{W_2}}) = X_{X_2}\), and \(\operatorname{image}(L\vert_{Z_{W_1 + W_2}}) = Z_{X_1 + X_2}\).

    By 1., this implies that there is an automorphism \(F \in \operatorname{Aut}(G)\) such that \(F(W_1) = X_1\) and \(F(W_2) = X_2\), where \(F(U) := \operatorname{image}(L\vert_{U})\) for all \(U \in S\).

    Hence \(G\) is distance-transitive, since \(W_1, W_2, X_1\) and \(X_2\) are arbitrary.

By 1. and 2., it follows that Grassmann graphs are distance-transitive in general.