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Let $X_{j}$ be elements of $SL(n, \mathbb{C})$ with entries ${X_{j}^{k,l}}${X_{j}^{k,l}}$. Express the polynomial$P: SL(n,\mathbb{C})^N \rightarrow \mathbb{C}$as a polynomial$Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C}$in variables${X_{j}^{k,l}}$Similar express the polynomial $$Q''(X_1, \cdots, X_n)$$ $$= \left( det(X_1) - 1 \right)\left( det(X_1)det(X_2) - 1 = \right) prod\limits_{i=1}^N \cdots\left( det(X_1)det(X_2) left( \cdots det(X_N) det(X_i) -1 \right)$$ in variables${X_{j}^{k,l}}$. You can then apply the well-known result for$\mathbb{C}^{N \cdot n}$n^2}$ to $Q' \cdot Q''$, and get what you want.

Edit due the comment: Note that we have both $$\{ Q' Q'' = 0 \} \subset \mathbb{C}^{N \cdot n^2}$$ is a connected subset, and that it is really a subset of $SL_n(\mathbb{C})^N$ $$\{ Q' Q'' = 0 \} \subset \{ Q'' =0 \} = SL_n(\mathbb{C})^N.$$

I am not sure, what topology you are working in, but it holds in any topology, the result original results for $\mathbb{C}^N$ holds in;)

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Simply write

Let $X_{j}$ be elements of $SL(n, \mathbb{C})$ with entries ${X_{j}^{k,l}}$

Express the polynomial $P: SL(N,\mathbb{C})^n SL(n,\mathbb{C})^N \rightarrow \mathbb{C}$ as a polynomial $Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C}$ times in variables ${X_{j}^{k,l}}$

Similar express the polynomial you get from $$Q''(X_1, \cdots, X_n)$$ $$= \left( det(X_1) - 1 \right)\left( det(X_1)det(X_2) - 1 \right) \cdots.$$ cdots\left( det(X_1)det(X_2) \cdots det(X_N) - 1 \right)$$in variables {X_{j}^{k,l}}. You can then simply apply the well-known result for \mathbb{C}^n, \mathbb{C}^{N \cdot n} to Q' \cdot Q'', and get what you want. I am not sure, what topology you are working in, but it holds in any topology, the result for \mathbb{C}^N holds in;) 3 corrected spelling Simple Simply write P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C} as a polynomial Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C} times the polynomial you get from$$ \left( det(X_1) - 1 \right)\left( det(X_1)det(X_2) - 1 \right) \cdots.

You can then simply apply the result for $\mathbb{C}^n$, and get what you want.

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