 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-3/6/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Number the vertices consecutively along the underlying chain. Orientation does not affect the quadratic Tits form of a quiver:

q_{Q} (n) = \sum_{i = 1}^{m} n_{i}^{2} - \sum_{i = 1}^{m - 1} n_{i} n_{i + 1} = \frac{1}{2} \sum_{i = 0}^{m} (n_{i + 1} - n_{i})^{2}, n_{0} = n_{m + 1} = 0.

n

has nonnegative integer coordinates and

q_{Q} (n) = 1

, the sum of integer squares on the right is

2

. There are therefore exactly two nonzero consecutive differences, each of absolute value one. Their sum is zero, so one is

+ 1

and the other

- 1

. Nonnegativity forces the

+ 1

to occur first. Thus the positive roots of type A are exactly the vectors with a single nonempty interval of ones and zeros elsewhere.

There is one such vector for every pair of endpoints

1 \leq a \leq b \leq m

# {positive roots of type A_{m}} = \frac{m ( m + 1 )}{2} .

For

A_{3}

, the full list is

(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (0, 1, 1), (1, 1, 1) .

The difference-of-coordinates proof includes

m = 1

and is independent of the chosen arrow orientation.

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