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I'm reading Landsberg's paper, which provide an introduction to geometric complexity theory. At chapter 2 of this paper, the author defined the following objects:

Let $W = \mathbb {C}^{n^2}$, $det_n \in S^nW$ be the determinant polynomial. And defined $$ Det_n := \overline{GL(W) \cdot [det_n]} \subseteq \mathbb {P}W $$

Where the bar denotes Zariski closure. I'm confused with the definition of $Det_n$. I will use the case $n=2$ to explain my confusion.

When $n=2$, we see $W =\mathbb {C}^4$ and $det_2 \in S^2W$. Let's denote a basis of $W$ be {$X_{11},X_{12},X_{21},X_{22}$}, then $det_2 = X_{11}X_{22}-X_{12}X_{21}$. Of course $GL(W)$ can acts on {$X_{11},X_{12},X_{21},X_{22}$} and their linear combination, and the result of such an action is still an element in $W$, and therefore an element in $\mathbb{P}V$, but there are product terms in $det_2$, so I can't understand how the result of such an action can be in $\mathbb{P}W$.

Thank for your help!

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    $\begingroup$ Where in the paper have you seen that $Det_n$ lives in $\mathbb{P}W$??? $\endgroup$
    – abx
    Commented Apr 22, 2020 at 9:43
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    $\begingroup$ @abx It's in the second paragraph of the Introduction chapter of this paper. $\endgroup$
    – Yi_Feng
    Commented Apr 22, 2020 at 10:38
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    $\begingroup$ No. There is no polynomials in that paragraph. Read it again. $\endgroup$
    – abx
    Commented Apr 22, 2020 at 12:01
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    $\begingroup$ @abx Thank you for your comments. So I think we can view $Det_n$ as a point in $\mathbb{P}S^nW$, and $GL(W)$ acts on it by acting on every symbols in every terms of the polynomial $Det_n$, do you think so? $\endgroup$
    – Yi_Feng
    Commented Apr 22, 2020 at 13:14
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    $\begingroup$ Yes. A representation of a group on a vector space extends in a natural way to its symmetric (or exterior) algebra. $\endgroup$
    – abx
    Commented Apr 22, 2020 at 13:59

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