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 Apr13 awarded Nice Question Apr9 answered analytical ability and logical reasoning Apr7 comment Tensor product in multilinear algebra This looks something like $(U\times V)^* \simeq U^* \otimes V^*$. Is this some alternative formulation of the tensor product? Or maybe it should be $\text{Hom}(U,V^*) \simeq U^* \otimes V^*$, because $Functions(X\times Y)$ is referring to the space of bilinear forms... Apr7 comment Tensor product in multilinear algebra @paulgarrett Hi, I have one more quick question :) It makes sense to me that the tensor product is a space whose dual is isomorphic to the space of bilinear forms (using some previous notation $(U\otimes V)^* \simeq \text{Hom}(U,V^*)$), and the link between elements of $U\times V$ and $U\otimes V$ is defined by a natural bilinear form $b$. But I've also seen that we should want our tensor product to satisfy a different property, something like $Functions(X\times Y) = Functions(X)\otimes Functions(Y)$. See the 2nd paragraph here: math.harvard.edu/archive/25b_spring_05/tensor.pdf Apr6 comment Tensor product in multilinear algebra @paulgarrett OK, thanks, this makes so much more sense now! I really like your explanation because it makes the link between the space of all bilinear forms over $U$ and $V$ and the space of all linear functionals on $U\otimes V$ extremely explicit. In other places I've heard that these two spaces should be "isomorphic", but at least in the finite-dimensional case (I don't know beyond that :) ) to be isomorphic they just need to be the same dimension, so this isn't saying much. But by your definition you must also exhibit the 'link' $b$ between the two spaces. You've made my day :) Apr6 comment Tensor product in multilinear algebra @paulgarrett Sorry about the $\oplus$ - I meant $\otimes$ actually - $B$ is a linear functional on $V\otimes W$, or more generally a linear map $V\otimes W \rightarrow X$. I'm still a little unsure now about how that composition works. If $\beta$ maps from $V\times W \rightarrow X$, how can we compose it with $B$ which maps from $V\otimes W \rightarrow X$? Wouldn't we need $B: X \rightarrow V\otimes W$ so that $b(x,y) = B(\beta(x,y)) \in V\otimes W$? Apr6 comment Tensor product in multilinear algebra @paulgarret However, you then go on to say that the link between $B$ and $\beta$ is that $B$ is the unique linear functional on $V\oplus W$ with the property that '$b = B \otimes \beta$'. I'm totally lost here. You've used $\otimes$ to signify a tensor product between two vector spaces, but $B$ and $\beta$ are linear and bilinear functionals respectively. I just don't know how to interpret what you mean by $B\otimes \beta$. Thanks Apr6 comment Tensor product in multilinear algebra @paulgarret I have a question about your characterization of a tensor product. You say the tensor product $V \otimes W$ is a vector space associated with a bilinear $b$ which maps elements of $V\times W$ to elements of $V \otimes W$. And you say that bilinear $b$ is special, because for any bilinear functional $\beta$ over $V$ and $W$ it allows us to find a unique, associated linear functional on $V \oplus W$. This is promising, as it suggests a link (possibly a homomorphism??) between the space of bilinear functionals on $V$ and $W$ and the space of linear functionals on $V\otimes W$. Mar25 comment Parameters in the Hamilton-Jacobi Equation Brilliant, thanks! Mar25 accepted Parameters in the Hamilton-Jacobi Equation Mar24 asked Parameters in the Hamilton-Jacobi Equation Mar19 awarded Nice Question Feb22 awarded Popular Question Dec15 awarded Caucus Nov17 awarded Good Question Oct5 awarded Popular Question Sep24 awarded Autobiographer Sep18 awarded Nice Answer Sep18 answered Is there any book/resource which explain the general idea of the proof of Fermat's last theorem? Sep18 answered Exponential random variable with mean 1/gamma