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`\left(\right)` produces too much space after `S^2`
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LSpice
LSpice
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The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$$U = S^2 \mathbb{C}^3 \oplus \mathbb{C}\cdot t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$$$ \operatorname{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2W \right),$$$$S^2{\left(\mathbb C^3 \cap W\right)} \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}\cdot t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \operatorname{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2{\left(\mathbb C^3 \cap W\right)} \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

deleted 2 characters in body
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Libli
Libli
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The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim V = 10$$\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2(W) \right),$$$$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2(W) \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2W \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.

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Libli
Libli
  • 7.3k
  • 25
  • 48

The answer is still no and a variant of my previous counter-example will provide a counter-example for this new question.

Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}.t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as: $$ \mathrm{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$

Note that $\dim U = 7$, $\dim V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have: $$S^2 \left(\mathbb{C}^3 \cap W \right) \subset \left(U \cap S^2(W) \right),$$ with $\dim \mathbb{C}^3 \cap W \geq 1$.