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Jul
13
comment $dx$ as a function
I also have no objections to this question - I wondered the same thing on learning this material. And yes, there's nothing ridiculous about $x'(x)$ - it's just a funny notation for the derivative of the function $f(x) = x$ (which is in particular the function of $f'(x) = 1$.)
Jun
16
answered Differential Equations Lectures or books from a theoretical perspective?
May
23
awarded  Critic
May
21
comment Uniqueness of Rectifying Coordinates: Question for Arnold's ODE Book
@Evgeny Yes, that was my initial intuition also, that perhaps the basis vectors were stretched. But I can't get the maths to work that way :)
May
19
asked Uniqueness of Rectifying Coordinates: Question for Arnold's ODE Book
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6
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13
awarded  Nice Question
Apr
9
answered analytical ability and logical reasoning
Apr
7
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...
Apr
7
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
Apr
6
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 :)
Apr
6
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 $b(x,y) = B(\beta(x,y)) \in V\otimes W$?
Apr
6
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
Apr
6
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 $V\otimes W$.
Mar
25
comment Parameters in the Hamilton-Jacobi Equation
Brilliant, thanks!
Mar
25
accepted Parameters in the Hamilton-Jacobi Equation
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24
asked Parameters in the Hamilton-Jacobi Equation
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19
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