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Apologies in advance that this is a somewhat soft question.

Let $k$ be an infinite field. Fix a dimension $d$ and let $v_1,\dots,v_r$, $w_1,\dots,w_r$ be two tuples of linearly independent vectors in $k^d$. (Thus, assume $r\leq d$.)

We have that the exterior products $v_1\wedge\dots\wedge v_r$ and $w_1\wedge\dots\wedge w_r$ are equal if and only if the $w_i$'s can be expressed in terms of the $v_i$'s via a matrix in the special linear group $SL(r,k)$. It follows that the map to the $r$th exterior power

$$\wedge^r:(k^d)^{\times r} \rightarrow \Lambda^rk^d$$

that carries

$$(v_1,\dots,v_r)\mapsto v_1\wedge\dots\wedge v_r$$

has fibers generically the size of $SL(r,k)$, which has dimension $r^2-1$ as an algebraic variety over $k$.

Now consider the same question with the symmetric and tensor powers in place of the exterior power. For the symmetric power, it seems to me we have $v_1\dots v_r = w_1\dots w_r$ (if and) only if there are elements $\alpha_1,\dots,\alpha_r\in k$ and a permutation $\sigma\in S_r$ (the symmetric group on $r$ elements) satisfying $\prod \alpha_i=1$, and $w_i=\alpha_iv_{\sigma(i)}$ for all $i$. I'll include an argument in a postscript, but if this is right, this means that the fibers of the map to the symmetric power

$$ \Pi^r: (k^d)^{\times r} \rightarrow S^rk^d$$

given by

$$ (v_1,\dots,v_r) \mapsto v_1\dots v_r $$

has fibers generically the size of the group $L\times S_r$, where $L$ is the set of matrices $\operatorname{diag}(\alpha_1,\dots,\alpha_r)$ with the $\alpha_i$'s as above, i.e. the diagonal matrices in $SL(r,k)$. This group has dimension $r-1$ as a $k$-variety.

Aside: I don't know a standard name for this map to the symmetric power. I am calling it $\Pi^r$ because if one interprets the symmetric algebra over $k^d$ as a polynomial ring, then it is just taking a product.

And then the more restrictive condition $v_1\otimes\dots\otimes v_r = w_1\otimes \dots\otimes w_r$ should only be met when $w_i = \alpha_iv_i$ for some $\alpha_1,\dots,\alpha_r\in k$ with $\prod\alpha_i=1$, as above, so the map

$$\otimes^r : (k^d)^{\times r} \rightarrow (k^d)^{\otimes r}$$

given by

$$(v_1,\dots,v_r)\mapsto v_1\otimes\dots\otimes v_r$$

has fibers generically bijecting just with $L = \{\operatorname{diag}(\alpha_1,\dots,\alpha_r)\}$, again, dimension $r-1$.

It is really bothering me that the fibers of $\Pi^r$ and $\otimes^r$ are the same dimension, and this dimension is so much smaller than the dimension of the fibers of $\wedge^r$. Can you offer any insight about why this is happening? (Or is my reasoning incorrect and it's not actually happening?)

Here is a little more about why it is bothering me. I normally think of the exterior algebra as quite a bit "smaller" than the symmetric algebra, but both are "massively smaller" than the tensor algebra. I mean this informally, but some form of it is true precisely: the dimension of $\Lambda^rk^d$ is $\binom{d}{r}$ which $\to 0$ as $r\to \infty$; that of $S^rk^d$ is $\binom{r+d-1}{d-1}$, which is polynomial in $r$ of degree $d-1$; and that of $(k^d)^{\otimes r}$ is $d^r$, which is obviously exponential in $r$.

In view of this, it's intuitive to me that the fibers of the map into the exterior power are the biggest (since smaller target perhaps forces bigger fibers), but it is confusing that the fibers of of the maps into the symmetric and tensor powers are the same size (well, at least same dimension), given how massively different in size the targets are.

As I write this, it occurs to me that, well, the source $(k^d)^{\times r}$ only has dimension $dr$, so $S^rk^d$ and $(k^d)^{\otimes r}$ both contain "plenty of room" to accommodate the source, while for some values of $r,d$, the dimension of $\Lambda^rk^d$ is much smaller than that of the source, forcing the fibers to embiggen. But this doesn't really feel explanatory to me, especially in view of the fact that, per the calculation above, none of the generic fiber sizes ever depend on $d$ (except inasmuch as the assumption $r\leq d$ is satisfied).

Can you offer any insight into what's going on?

Postscript: Here is my argument that $v_1\dots v_r = w_1\dots w_r$ (if and) only if there are elements $\alpha_1,\dots,\alpha_r\in k$ and a permutation $\sigma\in S_r$ (the symmetric group on $r$ elements) satisfying $\prod \alpha_i=1$, and $w_i=\alpha_iv_{\sigma(i)}$ for all $i$.

Extend $v_1,\dots,v_r$ to a basis of $k^d$. Interpret $S^\star k^d$ as the polynomial algebra over $k$ in $d$ indeterminates, but choose the indeterminates to be the basis just constructed; in particular, $v_1,\dots,v_r$ will be indeterminates, while $w_1,\dots,w_r$ will be linear forms.

Then $v_1\dots v_r$ is a monomial and $w_1\dots w_r$ is a homogeneous polynomial of degree $r$. Suppose we have $v_1\dots v_r = w_1\dots w_r$. A product of linear forms is not a monomial unless each linear form is a monomial. (One way to see this is to fix a monomial order; then the leading term and "trailing term" in an expanded product will be distinct unless each factor is a monomial.) Thus each $w_i$ is a coefficient times a single indeterminate. Furthermore, these indeterminates must coincide with $v_1,\dots,v_r$ in some order, or else $w_1\dots w_r$ and $v_1\dots v_r$ will be linearly independent. This establishes that there is a permutation $\sigma\in S_r$ and some numbers $\alpha_i\in k$ such that $w_i = \alpha_i v_{\sigma(i)}$ for all $i=1,\dots,r$. Thus the assumed equality $\prod v_i = \prod w_i$ becomes $\prod v_i = \prod \alpha_i \prod v_i$, and $\prod \alpha_i = 1$ follows.

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