Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 6 / a / ii

X

n

0 < q_{Q} (n) = 1 - dim Ext_{Q}^{1} (X, X) \leq 1.

The Tits form of a quiver takes integer values on integer vectors, so $q_{Q} (n) = 1$ and $X$ has no self-extensions.

Conversely, suppose

n

has nonnegative integer entries and

q_{Q} (n) = 1

. Choose a representation

M

of that dimension vector whose orbit has maximal dimension. Such an orbit exists because dimensions are integers bounded by

dim Rep_{Q} (n)

. If

M = U \oplus V

with nonzero

U, V

, a nonzero extension in either direction would, by part 5(c), produce a middle representation of the same dimension vector with larger orbit. Hence

Ext_{Q}^{1} (U, V) = Ext_{Q}^{1} (V, U) = 0

u = dim U

and

v = dim V

, the Ringel form identity then gives

1 = q_{Q} (u + v) = q_{Q} (u) + q_{Q} (v) + ⟨ u, v ⟩_{Q} + ⟨ v, u ⟩_{Q} = q_{Q} (u) + q_{Q} (v) + dim Hom_{Q} (U, V) + dim Hom_{Q} (V, U) \geq 2.

The last inequality uses positive integral values of

q_{Q}

at both nonzero vectors. This contradiction makes

M

indecomposable. Thus the indecomposable dimension vectors are exactly the positive roots $q_{Q} (n) = 1$ .

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