Tensor can be thought of as generalization vectors but tensor is described in many ways sometimes an array of numbers.
What is the most common and appropriate definition of tensor ?
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2$\begingroup$ They are indeed (multidimensional) arrays of numbers, but Wikipedia associates much more semantics, from differential geometry, to the definition: en.wikipedia.org/wiki/Tensor. $\endgroup$– user65203Commented Jul 24, 2019 at 8:26
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2$\begingroup$ A tensor space is a quotient space w.r.t. to the smallest subspace which contains the defining equations of the tensors. There is no way to get around it. $\endgroup$– WuestenfuxCommented Jul 24, 2019 at 8:30
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1$\begingroup$ I used to resist the definition of a tensor as a mere multidimensional array of numbers. But then I realized this definition is used in Golub and Van Loan (Matrix Computations). So now I'm ok with that use of the term. $\endgroup$– littleOCommented Jul 24, 2019 at 8:42
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1$\begingroup$ If $V$ is a finite dimensional $K$-vector with basis $(b_j)$ that is $V = \sum_{j=1}^J b_j K$ and $W = \sum_{i=1}^I a_i K$ then $V \otimes_K W$ is the $JI$-dimensional $K$-vector space $\sum_{j=1}^J \sum_{i=1}^I (b_j \otimes_K a_i) K$. Here $b_j \otimes_K a_i$ is a just a symbol denoting a pair. But the obtained vector space doesn't depend on the chosen basis, so $\otimes_K$ obeys some bilinearity laws. Then the same definition extends to $K$-algebras, look at the algebra $End_K(V) \otimes_K End_K(W)$, how it acts on $V \otimes_K W$, what does the tensor product of number fields look like.. $\endgroup$– reunsCommented Jul 24, 2019 at 16:06
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1$\begingroup$ @Wuestenfux: There are plenty of ways to get around it! The definition you give is the most standard one in maths today, but there are several other constructions that are just as “correct” (i.e. they give an isomorphic result), some of which have been more standard in the past and/or in other fields (e.g. physics). $\endgroup$– Peter LeFanu LumsdaineCommented Jul 24, 2019 at 16:34
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1 Answer
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I favor the abstract definition of a tensor space as a quotient vector space.
We define the tensor product of vector speces $V$ and $W$ over a common base field as the quotient vector space: $$ V \otimes W := F(V\times W)/\sim$$ where $F(Z)$ is a free vector space generated by elements of set $Z$, and $\sim$ is the minimimal equivalence relation such that
- $(v,w)+(v',w) \sim (v+v',w)$ and $(v,w)+(v,w') \sim (v,w+w')$
- $(\lambda v, w) \sim \lambda(v,w) \sim (v,\lambda w)$
This definition can be generalized to define a tensor product of arbitrary number of vector spaces. A tensor is an element of a tensor space. We define $v\otimes w := [(v,w)]_\sim$
I like this definition because from it immediately follow the important arithmetic properties of tensors:
- $ (v\otimes w)+(v'\otimes w) = (v+v')\otimes w $ and $ (v\otimes w)+(v\otimes w') = v\otimes (w+w') $
- $ (\lambda v)\otimes w = \lambda(v\otimes w) = v \otimes (\lambda w)$
It also shows that a tensor product is uniquely defined and independent of the bases of the vector spaces. But given specific bases of $V$ and $W$, we can easily construct an isomorphism between the abstract tensor product and the array of numbers. We just need to remember that this isomorphism is basis-dependent.
