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Is it true that any element of $\mathbb{R}{\otimes}_{\mathbb{Q}}\mathbb{R}$ is symmetrical?


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What do you mean by symmetrical? that it's of the form $x\otimes x$? Then no, for example $1\otimes 2$ is there. – Dennis Gulko Feb 17 '12 at 21:41

If by symmetrical you mean invariant under the map $x\otimes y\mapsto y\otimes x$, then no. If you take a basis $\{\alpha_i:i\in I\}$ for $\mathbb{R}$ over $\mathbb{Q}$ and distinct elements $\alpha_1$ and $\alpha_2$ of this basis, then $\alpha_1\otimes\alpha_2\neq\alpha_2\otimes\alpha_1$. This is because every element of $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$ is uniquely of the form $\sum_{i\in I}\alpha_i\otimes\beta_i$ for real numbers $\beta_i$, all but finitely many of which are equal to zero.

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