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I am looking at isometries of the two-torus with the flat metric. From a local perspective, $T^2 = S^1 \times S^1$ has coordinates $(x,y)$, with metric: $$ds^2 = dx^2 + dy^2$$ The difference between $T^2$ and $\mathbb{R}^2$ is in the topology, and comes from imposing the periodicity of the coordinates $x \sim x+1$ and $y \sim y+1$.

I know that $\mathbb{R}^2$ has three Killing vectors: $\{ \partial_x,\partial_y, x\partial_y - y\partial_x \}$, which are simply the translations, and rotation of the plane. These are also local Killing vectors for the torus, but on the other hand I know that $T^2$ has only two global Killing vectors, the translations.

How, explicitly, do I see that the rotation Killing vector is not a global Killing vector for the two-torus?

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    Is the function $x$ well-defined on the torus? – 2017-02-07
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    No, it is only defined locally. So I should be able to compare the Killing vector at two points, $p = (x,y)$, and $p' = (x+1,y)$. In the manifold, $p = p'$, but the Killing vectors at these two points will be different, so the Killing vector is not globally defined on the manifold? – 2017-02-07
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    Think about it geometrically. Does rotation preserve the lattice? – 2017-02-07
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    @MarkB: Yep, that's the idea. – 2017-02-07

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Based on the comments by Jack Lee, the following is an easy way to see that the Killing vector is not globally defined:

On the manifold, the point $p=(x,y)$ is identified with the point $p'=(x+1,y)$. In order for the Killing vector $k=x\partial_y -y\partial_x$ to be globally defined, it should be the same at these two points. But we note that the Killing vector at $p'$ is given by $k(p')=x\partial_y -y\partial_x + \partial_y =k(p) + \partial_y $. It follows then that $k$ is not a globally defined Killing vector.