There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous equality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ is $\ d\binom{r+k-1}{r}$.

Finally, if $E$ is stable and the characteristic is 0, $Sym^k(E)$ is semi-stable: this follows from the Narasimhan-Seshadri descriptionn of (semi-) stable bundles in terms of unitary representations. It is easy to find examples where $Sym^k(E)$ is not stable; there are also examples in characteristic $p$ where it is not semi-stable.

There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous equality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ is $\ d\binom{r+k-1}{r}$.

Finally, if $E$ is stable and the characteristic is 0, $Sym^k(E)$ is semi-stable: this follows from the Narasimhan-Seshadri description of (semi-) stable bundles in terms of unitary representations. It is easy to find examples where $Sym^k(E)$ is not stable; there are also examples in characteristic $p$ where it is not semi-stable.

There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous inequality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ is $\ d\binom{r+k-1}{r}$.

Finally, if $E$ is stable and the characteristic is 0, $Sym^k(E)$ is semi-stable: this follows from the Narasimhan-Seshadri descriptionn of (semi-) stable bundles in terms of unitary representations. It is easy to find examples where $Sym^k(E)$ is not stable; there are also examples in characteristic $p$ where it is not semi-stable.

There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous equality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ is $\ d\binom{r+k-1}{r}$.

Finally, if $E$ is stable and the characteristic is 0, $Sym^k(E)$ is semi-stable: this follows from the Narasimhan-Seshadri descriptionn of (semi-) stable bundles in terms of unitary representations. It is easy to find examples where $Sym^k(E)$ is not stable; there are also examples in characteristic $p$ where it is not semi-stable.