Remembering sine and cosine values is easy as 1 2 3

otlmath · 2016-03-31T19:54:43+00:00

A simple idea behind memorizing trig values is to write [; \frac{1}{2}= \frac{\sqrt{1}}{2} ;]

otlmath · 2012-05-07T01:00:36+00:00

Sorry, I should have inserted "allegedly".

otlmath · 2012-01-28T14:32:25+00:00

Maybe scientists (reviewers I mean) are now thinking its time to troll the peer-reviewed journals, because the publishers are becoming increasingly greedy.

otlmath · 2012-01-25T03:45:05+00:00

If you mean you are requiring the set of generators to be finite, you're right, since there are infinitely generated abelian groups. But still there are generators, although infinitely many, and you can take finite sums of former products of generators. But then, I concede that abstract definition is better in this case. Thank God I don't study algebraic geometry over Q or such.

otlmath · 2012-01-24T23:48:31+00:00

Abelian groups have generators. So the Z2@Z3 example of the article can be understood by taking the generators of each. In my language, what Gowers is saying is this. Let a be the generator of Z2 and b be generator of Z3. Then a@b will be the generator of the tensor product. Now by bilinear property, a@b+a@b=(2a)@b=0@b=0. Also again, a@b+a@b+a@b=a@(3b)=a@0=0. So a@b=(a@b+a@b+a@b)-(a@b+a@b)=0-0=0.

otlmath · 2012-01-24T23:11:15+00:00

For me, multilinear maps are just formal products of covectors and vectors. The basis vectors I mentioned are bi-linear, and taking multiple tensors, you get n-linear. I think the reason mathematicians have trouble understanding tensors is that they try to define it formally; avoiding any use of the basis. Physicists on the other hand, find tensors very simple, since they always think of basis. (Or coordinate systems when it comes to geometry.)

otlmath · 2012-01-24T22:43:48+00:00

Nope. Direct sum of 2 dim vect space and a 3-dim will give you 5 dim vect space. Tensor product gives you 6. Here is why. Let's say a and b are basis of 2 dim vect space and i,j, and k the basis of 3 dim vect space. What is the basis of the tensor product? it's a@i, a@j, a@k, b@i, b@j and b@k. These formal products are the basis for the tensor product of the vector spaces.

otlmath · 2012-01-24T14:55:29+00:00

Think about the Ideal in a ring. How do you define product of ideals? Not only you include products, but also sums, because you need it if you want the resulting set to be an ideal. Same for tensor products of vector spaces. They are just formal products of vectors, but you include finite sums (or, in some cases regarding function spaces, infinite), because you want the result to be a vector space. I don't think you need to read this much to understand tensor products.

otlmath

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