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Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 0\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Update

Using Tarski-Seidenberg (a proposed by Robert in the comments) it is relatively easy to see that $C^\circ$ is indeed semi-algebraic $-$ but for a more general notion of semi-algebraic as intended in my question (see my answer below). Namely, $C^\circ$ can be written as the intersection and union of algebraic sets, where in $(*)$ I only want to use intersection. I suspect that one can get rid on the unions by using that $C^\circ$ is convex.

Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 0\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 1\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Update

Using Tarski-Seidenberg (a proposed by Robert in the comments) it is relatively easy to see that $C^\circ$ is indeed semi-algebraic $-$ but for a more general notion of semi-algebraic as intended in my question (see my answer below). Namely, $C^\circ$ can be written as the intersection and union of algebraic sets, where in $(*)$ I only want to use intersection. I suspect that one can get rid on the unions by using that $C^\circ$ is convex.

Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 0\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Update

Using Tarski-Seidenberg (a proposed by Robert in the comments) it is relatively easy to see that $C^\circ$ is indeed semi-algebraic $-$ but for a more general notion of semi-algebraic as intended in my question (see my answer below). Namely, $C^\circ$ can be written as the intersection and union of algebraic sets, where in $(*)$ I only want to use intersection. I suspect that one can get rid on the unions by using that $C^\circ$ is convex.

Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 1\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Secondary questions: If the answer to my question is No, what can be said about the class of semi-algebraic convex bodies for which this is true? If Yes, how would one compute a representation as in $(*)$ for $C^\circ$? I can imagine that there exists quite some literature in this direction (real algebraic geoemtry? algebraic convex geometry? semi-algebraic geometry?), but I could not find anything specific and I am happy about any keyword or reference.

Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as

$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 1\}$$

with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means

$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$

Update

Using Tarski-Seidenberg (a proposed by Robert in the comments) it is relatively easy to see that $C^\circ$ is indeed semi-algebraic $-$ but for a more general notion of semi-algebraic as intended in my question (see my answer below). Namely, $C^\circ$ can be written as the intersection and union of algebraic sets, where in $(*)$ I only want to use intersection. I suspect that one can get rid on the unions by using that $C^\circ$ is convex.