Linear algebraic group of $GL(V)$ is independent of choice of basis

Question

This is an easy question, but I am confused now. Please see the definition in this book

In the last line. If we have $f(g)=0$, where $f(y)=p(x_{11}(y),...,x_{nn}(y))$, then after choosing different basis for $V$, do we still have $f(P^{-1}gP)=0$? Note that $p\in\mathbb{C}[(x_{11},x_{12}...,x_{nn}]$ , I am confused why the definition is independent of choice of basis.

Answer 1

The group $GL(V)$ is defined as the variety defined by the zero set of the polynomial $(x,y) \mapsto \det(x) * y-1$, where $x \in End(V)$ and $y \in \mathbb{C}$.

The determinant is invariant under basis transformation.

Edit: I didn't address the question in its new form. Changing the basis gives a conjugated embedding. So they are not the same groups any more, but conjugated ones. So considering an abstract linear algebraic group and a subgroup of $GL(V)$ is slightly different, but not essential.

Answer 2

Let's say you have a basis $\mathcal{V}=\{v_1,\ldots,v_n\}$ of $V$ and let $\mu_\mathcal{V}:GL(V)\to GL(n,\mathbb{C})$ be the isomorphism $\mu$ defined in the book. Since what $\mu$ actually maps to depends on the basis, I don't drop the subscript $\mathcal{V}$ here. Then a subgroup $G\subseteq GL(V)$ is a linear algebraic subgroup of $G$ if $\color{red}{\mu_\mathcal{V}(G)}$ is a set of the form $$ \{g\in GL(n,\mathbb{C}): f(g) = 0 \text{ for all } f\in\mathcal{A}\}\tag{1} $$ for some set of polynomials $\mathcal{A}$ in $M_n(\mathbb{C})$. Note that on the displayed page, the symbol $G$ in definition 1.4.1 and in the subsequent paragraphs mean different things, but if we are saying that $G$ is a linear algebraic subgroup of $GL(V)$, the set in $(1)$ should be $\mu_\mathcal{V}(G)$, not $G$.

Now, if you change the basis to some $\mathcal{U}=\{u_1,\ldots,u_n\}$ and the change-of-basis matrix from $\mathcal{V}$ to $\mathcal{U}$ is $P$, then $$ \mu_\mathcal{U}(G) = P\mu_\mathcal{V}(G)P^{-1}. $$ It follows that $$ \color{red}{\mu_\mathcal{U}(G)} = \{\widetilde{f}\in GL(n,\mathbb{C}): \widetilde{f}(g) = 0 \text{ for all } \widetilde{f}\in \widetilde{\mathcal{A}}\}\tag{2} $$ where $$ \widetilde{\mathcal{A}} = \left\{\widetilde{f}: \widetilde{f}(g) = f(P^{-1}gP) \text{ for all } g\in M_n(\mathbb{C})\right\}. $$

Now, by saying that the definition of linear algebraic group does not depend on the choice of basis, we mean, upon a change of basis, $\mu_\mathcal{U}(G)$ is the zero set of some set of polynomials. Note that this new set of polynomials ($\widetilde{\mathcal{A}}$) are in general different from the old one ($\mathcal{A}$), and $\mu_\mathcal{V}(G),\ \mu_\mathcal{U}(G)$ are different but isomorphic subgroups of $GL(n,\mathbb{C})$.