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0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not necessarily expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. In other words, the infinitely many inequation relations satisfied by elements of $X_n$ lie on a real variety, meaning they probably have some sort of nice structure. Someone who knows more than I do should try to answer the rest of the question.

0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $2n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not necessarily expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. In other words, the infinitely many inequation relations satisfied by elements of $X_n$ lie on a real variety themselves, meaning they probably have some sort of nice structure. Someone who knows more than I do should try to answer the rest of the question.

0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. Someone who knows more than I do should try to answer the rest of the question.

0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not necessarily expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. In other words, the infinitely many inequation relations satisfied by elements of $X_n$ lie on a real variety, meaning they probably have some sort of nice structure. Someone who knows more than I do should try to answer the rest of the question.

0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not expect to be a variety. It might, however, be a kind of quasiprojective real variety. Someone who knows more than I do should try to answer the rest of the question.

0. Under most reasonable notions of dimension, the (real) dimension would be $2n$. For a polynomial up to scaling is the same as an element of $\mathbb{C}^n$ (i.e. a collection of $n$ roots). We can mod out by scaling by a positive real factor, giving something of real dimension $2n-1$. When doing this modding out, the fibre of each element of the quotient contains at most $n$ (hence a finite number) of polynomials which are not in $X_n$, implying that the real (topological) dimension should still be $n$.

As for whether it's a variety, you're looking at the intersection of infinitely many Zariski closed sets defined by $\sum_{i=0}^n w_i z^i$ for each $z \in S^1$, which I would not expect to be a variety. It is, however, a definable real set (in the sense of model theory), which would imply that it's at least almost a kind of quasiprojective real variety. Someone who knows more than I do should try to answer the rest of the question.