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Let $\mathrm{E}$ be a vector bundle of rank $2$ with $c_1(\mathrm{E})=0$ on $\mathbf{P}^4$. Then the Chern character of $\mathrm{E}$ can be written as $$\mathrm{ch}(\mathrm{E})=\sum_{m=0}^\infty \frac{2(-1)^m}{(2m)!}c_2(\mathrm{E})^m=2-c_2+\frac{1}{12}c_2^2+\ldots.$$ The degree $4$ part of the Todd class of $\mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $\mathrm{E}$ is $$\chi(\mathrm{E})=2\chi(\mathrm{O}_{\mathbf{P}^4})+\frac{1}{12}\int_{\mathbf{P}^4} \big(-35c_2 h^4+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $\mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$. The case $c_2=3$ was ruled out by Barth and Elencwajg; the latter is known to frequent this site, no doubt he has much more to say.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $\mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $\mathrm{T}_{\mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $\mathrm{E}$ on $\mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)

Let $\mathrm{E}$ be a vector bundle of rank $2$ with $c_1(\mathrm{E})=0$ on $\mathbf{P}^4$. Then the Chern character of $\mathrm{E}$ can be written as $$\mathrm{ch}(\mathrm{E})=\sum_{m=0}^\infty \frac{2(-1)^m}{(2m)!}c_2(\mathrm{E})^m=2-c_2+\frac{1}{12}c_2^2+\ldots.$$ The degree $4$ part of the Todd class of $\mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $\mathrm{E}$ is $$\chi(\mathrm{E})=2\chi(\mathrm{O}_{\mathbf{P}^4})+\frac{1}{12}\int_{\mathbf{P}^4} \big(-35c_2 h^2+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $\mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$. The case $c_2=3$ was ruled out by Barth and Elencwajg; the latter is known to frequent this site, no doubt he has much more to say.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $\mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $\mathrm{T}_{\mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $\mathrm{E}$ on $\mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)

Let $\mathrm{E}$ be a vector bundle of rank $2$ with $c_1(\mathrm{E})=0$ on $\mathbf{P}^4$. Then the Chern character of $\mathrm{E}$ can be written as $$\mathrm{ch}(\mathrm{E})=\sum_{m=0}^\infty \frac{2(-1)^m}{(2m)!}c_2(\mathrm{E})^m=2-c_2+\frac{1}{12}c_2^2+\ldots.$$ The degree $4$ part of the Todd class of $\mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $\mathrm{E}$ is $$\chi(\mathrm{E})=2\chi(\mathrm{O}_{\mathbf{P}^4})+\frac{1}{12}\int_{\mathbf{P}^4} \big(-35c_2 h^4+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $\mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $\mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $\mathrm{T}_{\mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $\mathrm{E}$ on $\mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)

Let $\mathrm{E}$ be a vector bundle of rank $2$ with $c_1(\mathrm{E})=0$ on $\mathbf{P}^4$. Then the Chern character of $\mathrm{E}$ can be written as $$\mathrm{ch}(\mathrm{E})=\sum_{m=0}^\infty \frac{2(-1)^m}{(2m)!}c_2(\mathrm{E})^m=2-c_2+\frac{1}{12}c_2^2+\ldots.$$ The degree $4$ part of the Todd class of $\mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $\mathrm{E}$ is $$\chi(\mathrm{E})=2\chi(\mathrm{O}_{\mathbf{P}^4})+\frac{1}{12}\int_{\mathbf{P}^4} \big(-35c_2 h^4+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $\mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$. The case $c_2=3$ was ruled out by Barth and Elencwajg; the latter is known to frequent this site, no doubt he has much more to say.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $\mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $\mathrm{T}_{\mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $\mathrm{E}$ on $\mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)

This follows from the `Schwarzenberger condition' $c_2 (c_2+1-3c_1-2c_1^2)\equiv 0 \ \mathrm{mod} \ 12$.

Let $\mathrm{E}$ be a vector bundle of rank $2$ with $c_1(\mathrm{E})=0$ on $\mathbf{P}^4$. Then the Chern character of $\mathrm{E}$ can be written as $$\mathrm{ch}(\mathrm{E})=\sum_{m=0}^\infty \frac{2(-1)^m}{(2m)!}c_2(\mathrm{E})^m=2-c_2+\frac{1}{12}c_2^2+\ldots.$$ The degree $4$ part of the Todd class of $\mathbf{P}^4$ is $35h^2/12$, and the Hirzebruch-Riemann-Roch theorem says that the holomorphic Euler characteristic of $\mathrm{E}$ is $$\chi(\mathrm{E})=2\chi(\mathrm{O}_{\mathbf{P}^4})+\frac{1}{12}\int_{\mathbf{P}^4} \big(-35c_2 h^4+c_2^2).$$ Integrality requires that (identifying Chern classes with integers) $-35 c_2+c_2^2$ is divisible by $12$; in other words, that $c_2+c_2^2$ is divisible by $12$. In particular, there are no such $\mathrm{E}$ with $c_1=0$ and $c_2=1$ or $2$.

Schwarzenberger noticed that the Hirzebruch-Riemann-Roch theorem imposes divisibility constraints on the Chern classes, and what I explain above is merely a special case. For rank $2$ bundles on $\mathbf{P}^3$ you get the (probably most known) constraint that $c_1 c_2$ must be even. You can use this to show that $\mathrm{T}_{\mathbf{P}^2}$ cannot be the restriction of a rank $2$ bundle $\mathrm{E}$ on $\mathbf{P}^3$.

(A comprehensive discussion of such questions can be found in the standard reference `Vector Bundles on Complex Projective Spaces' by Okonek-Schneider-Spindler.)