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How are R^2 and R^4 isomorphic as VS's over Q?

by joeblow
Tags: tensor product
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joeblow
#1
Mar24-11, 08:56 PM
P: 71
I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R2 and the second is isomorphic to R4 over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)
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lavinia
#2
Mar25-11, 04:59 AM
Sci Advisor
P: 1,716
Quote Quote by joeblow View Post
I came across a problem asserting that the C-tensor product of C and C and the R-tensor product of C and C are isomorphic as Q-modules. How does this begin to make sense as they do not have the same dimension over R? The first is isomorphic to R2 and the second is isomorphic to R4 over R.

Also, I'm struggling with tensors. Does anyone have a good source I should check out? (I've tried many)
Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.

The tensor product is a way to extend the field of scalars of a vector space. A real vector space of dimension n (n can be infinite) becomes a complex vector space of dimension n when tensored with C. It becomes a real vector space of twice the dimension over R.

If one views R as a 1 dimensional vector space over itself with single basis vector v, the all elements are of the form rV for real numbers,r. Over C the basis is also v and the resulting vector space is 1 dimensional over C. But over R the basis is V and iV and so is 2 dimensional over R.
Landau
#3
Mar25-11, 03:55 PM
Sci Advisor
P: 905
Quote Quote by lavinia View Post
Over Q these vector spaces are infinite dimensional. They will be isomorphic if the cardinality of their bases is the same - which I think is the cardinality of the Continuum.
Indeed; we discussed this here.


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