Since {\bf P}^n http://latex.mathoverflow.net/png?%7B%5Cbf%20P%7D%5En is the quotient of {\bf A}^{n+1} - 0 http://latex.mathoverflow.net/png?%7B%5Cbf%20A%7D%5E%7Bn%2B1%7D%20%2D%200 by the action of {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm, the tangent bundle of {\bf P}^n http://latex.mathoverflow.net/png?%7B%5Cbf%20P%7D%5En is the quotient of the tangent bundle of {\bf A}^{n+1} - 0 http://latex.mathoverflow.net/png?%7B%5Cbf%20A%7D%5E%7Bn%2B1%7D%20%2D%200 by the action of the tangent bundle of {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm:

T {\bf P}^n = T ({\bf A}^{n+1} - 0) / T {\bf G}\sb m http://latex.mathoverflow.net/png?T%20%7B%5Cbf%20P%7D%5En%20%3D%20T%20%28%7B%5Cbf%20A%7D%5E%7Bn%2B1%7D%20%2D%200%29%20%2F%20T%20%7B%5Cbf%20G%7D%5Fm .

As a group, the tangent bundle of {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm is the product of {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm and a 1-dimensional vector space V. Therefore we can take the quotient of everything on the right side above by {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm. Note that T({\bf A}^{n+1}-0) http://latex.mathoverflow.net/png?T%20%28%20%7B%5Cbf%20A%7D%5E%7Bn%2B1%7D%20%2D%200%20%29%0A is the product of {\bf A}^{n+1} - 0 http://latex.mathoverflow.net/png?%7B%5Cbf%20A%7D%5E%7Bn%2B1%7D%20%2D%200 with the direct sum of (n+1) copies of the weight one representation of {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm. Therefore its quotient by {\bf G}\sb m http://latex.mathoverflow.net/png?%7B%5Cbf%20G%7D%5Fm is \mathcal{O}\sb {{\bf P}^n}(1)^{n+1} http://latex.mathoverflow.net/png?%5Cmathcal%7BO%7D%5F%7B%7B%5Cbf%20P%7D%5En%7D%281%29%5E%7Bn%2B1%7D%0A. We get

T {\bf P}^n = \mathcal{O}\sb {{\bf P}^n}(1)^{n+1} / V http://latex.mathoverflow.net/png?T%20%7B%5Cbf%20P%7D%5En%20%3D%20%5Cmathcal%7BO%7D%5F%7B%7B%5Cbf%20P%7D%5En%7D%281%29%5E%7Bn%2B1%7D%20%2F%20V%0A .

A (linear) action of a one-dimensional vector space on a vector bundle is the same thing as an exact sequence, so this gives the Euler sequence.

Since ${\bf P}^n$ is the quotient of ${\bf A}^{n+1} - 0$ by the action of ${\bf G}_m$, the tangent bundle of ${\bf P}^n$ is the quotient of the tangent bundle of ${\bf A}^{n+1} - 0$ by the action of the tangent bundle of ${\bf G}_m$:

$T {\bf P}^n = T ({\bf A}^{n+1} - 0) / T {\bf G}_m$ .

As a group, the tangent bundle of ${\bf G}_m$ is the product of ${\bf G}_m$ and a 1-dimensional vector space V. Therefore we can take the quotient of everything on the right side above by ${\bf G}_m$. Note that $T({\bf A}^{n+1}-0)$ is the product of ${\bf A}^{n+1} - 0$ with the direct sum of (n+1) copies of the weight one representation of ${\bf G}_m$. Therefore its quotient by ${\bf G}_m$ is $\mathcal{O}_{{\bf P}^n}(1)^{n+1}$. We get

$T {\bf P}^n = \mathcal{O}_{{\bf P}^n}(1)^{n+1} / V$ .

A (linear) action of a one-dimensional vector space on a vector bundle is the same thing as an exact sequence, so this gives the Euler sequence.