The dimension count is not right. 64 dimensions of automorphisms, 56 dimensions of forms. There are only finitely many orbits over $\C$.

I think this definition makes $G_2$ is a classical group: Look at $GL_7$ acting on the third exterior power of its standard representation; the connected component of a generic point stabilizer is $G_2$.

The answer is no, for the sixth symmetric power in characteristic zero. But I don't know if there is an easy proof. See "Symmetric powers and a problem of Kollar and Larsen," by Guralnick and Thiep.

BCnrd mentions superrigidity, and it seems likely it follows from that -- I don't see why the isomorphism needs to preserve arithmeticity. Such an isomorphism gives a representation $f: \Gamma \rightarrow (D')_v^{\times}$, where $\Gamma$ is the $T$-arithmetic subgroup of $D$ and $v$ a place. I think if one chooses $T$ large enough the key assumptions of superrigidity are satisfied (rank bigger than 2, big enough image); it then asserts that $f$ is ``of algebraic origin,'' and this should be enough. There are many little details to check of course.

One can also use Stone-Weierstrass directly: it implies that character of $W$ can be well-approximated by a polynomial in the matrix entries of $V$ and their conjugates. So $\chi_W$ is not orthogonal to some matrix entry for the representation $V^{\otimes k} \otimes (V^*)^{\otimes l}$. This implies that there is a nonzero homomorphism from $W$ to $V^{\otimes k} \otimes (V^*)^{\otimes l})$.

Take $R$ to be a field containing $\mathbb{C}$, and let $R$ act on itself by multiplication.

Take the conjugacy class of your (not necessarily semisimple) matrix inside $GL(n, \mathbb{C})$, and take its Zariski closure (or its closure in the usual topology). This contains a unique semisimple conjugacy class.

The limit as $q \rightarrow 1$ along the real axis, say, is certainly $0$, since $f$ is a cusp form and $q=1$ corresponds to $z=0$, which is a cusp. This is easy to see numerically for small conductor; the convergence to zero is staggeringly fast. Incidentally, Euler was already aware of this phenomenon in the related case $\sum_n (-1)^n q^{n^2}$ as $q \rightarrow 1$.

Correction to previous comment: The cover $C' \rightarrow C$ defined by $5$-torsion in relative Jacobian isn't Galois; indeed, we are only interested in a "certain piece" of its cohomology, but I'm not sure how to find this "piece" without passing to the "Galois closure"... And the Galois closure of $C' \rightarrow C$ is a cover of monstrous degree indeed.

In that question, we were working with mod $5$ coefficients. Also, $125$ should have been $625$. Finally, it is not "equivalent": you need only a piece of the cohomology of the cover. Specifically, if $f : X \rightarrow C$, let $C'$ be the $5$-torsion in the relative Jacobian; its Galois group is a subgroup of $\mathrm{GSp}(\mathbb{F}_5)$, and you need roughly the piece of the cohomology of $C'$ through which $G$ acts by the standard representation.

You should multiply by $\pi$ (that's the constant of proportionality). Then it looks a lot more like $\sqrt{\zeta(2)}$.

Two things: The density of ideals (as measured by norms) is not the same as the class number; rather, it's proportional to $h(D)/\sqrt{D}$ in the imaginary quadratic case. Secondly, the phrase "with large prime factors" makes a significant difference. I should say I didn't test this experimentally, but playing with it on paper it seems like the answers are as Scholz mentioned.

ideal density is residue at $1$ of $\zeta_k$; if you take geometric mean of this over fields $k$, I think you get quantities as stated. No idea if this is what he meant.