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A tensor of type $(r,s)$ on a vector space $V$ is a $C$-valued function $T$ on $V×V×...×V×W×W×...×W$ (there are $r$ $V$'s and $s$ $W$'s in which $W$ is the dual space of $V$) which is linear in each argument. We take $(0, 0)$-tensors to be scalars, as a matter of convention. The interpretations of $(r,0)$-tensors are trivial, since they are definitions of multilinear functionals (as a special case $(1,0)$-tensor interpreted as covector (elements of dual space)). We can interpret $(1,1)$ tensors as follows: $A(v,f ) ≡ f (Av)$. Say we have a linear operator $R$; then we can turn $R$ into a second rank tensor $T$ by $T(v,w) ≡ v · Rw$ where $·$ denotes the usual dot product of vectors. If we compute the components of $T$ we find that the components of the tensor $T$ are the same as the components of the linear operator $R$. Ok. Everything is good. But I cant understand interpretations of other $(r,s)$-tensors. For example I found in Wikipedia $(0,1)$-tensor interpreted as a vector or $(0,2)$ as a bivector and in general $(0,s)$ tensor as $s$-vector tensor; or $(2,1)$ tensor as cross product and so on. I want you to show how the tensors in general interpreted. Is it possible for you to show these interpretations like as I did for $(1,1)$-tensor ?

Alex R.
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  • Cross-posted to http://physics.stackexchange.com/q/128896/2451 – Qmechanic Jul 31 '14 at 11:56
  • Regarding why a $(0,1)$ tensor can be considered a vector, that is because (for finite-dimensional vector spaces) any vector space is isomorphic to its double dual vector space. And it seems like, using your definition, a $(0,1)$ tensor corresponds to an element of the double dual space. See the question an answers for much better explanation than what I am giving here: https://math.stackexchange.com/a/3229608/327486 – Chill2Macht Feb 26 '20 at 01:30
  • As the accepted answer below says, bivectors correspond only to alternating $(0,2)$ tensor, and general $n$-vectors are used to refer only to alternating tensors. Likewise, a $(2,1)$ tensor is a cross product only if it is also an alternating tensor. Being a $(2,1)$ tensor in general says that you are bilinear and take two vectors as input and give one vector as output. A cross product is an alternating bilinear function which takes two vectors as input and returns one vector as output. A lot of confusion will follow from general tensors being conflated with alternating tensors. – Chill2Macht Feb 26 '20 at 01:33

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Any alternating $(r,s)$ tensor has a corresponding map that goes $\Lambda^r V \to \Lambda^s V$. Suppose $R \in \Lambda^r V$ and $\Sigma \in \Lambda^s V^*$. Then define $\underline T:\Lambda^r V \to \Lambda^s V$ such that

$$T(R, \Sigma) = \Sigma[ \underline T(R)]$$

The uniqueness of $\underline T$ can be proved by taking a "gradient" with respect to the vector space of $\Lambda^s V^*$.

Geometrically, $\underline T$ maps an $r$-vector (which corresponds to an $r$-dimensional subspace) to an $s$-vector, and the $s$-covector $\Sigma$ allows us to extract the components of $\underline T(R)$.

Muphrid
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  • Thank you for your answer. I am not familiar with "alternating tensor" concept and its relation to tensor concept. Is it possible for you to explain more for me. –  Aug 01 '14 at 07:11
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    An alternating tensor is merely antisymmetric on interchange of adjacent arguments. For instance, if you had a $(2,0)$ tensor $T(u,v)$, it would be alternating if $T(u,v) = -T(v,u)$. When a tensor is alternating in this way, the vector and covector arguments can be reinterpreted as forming elements of the exterior algebra, corresponding to subspaces (or weighted sums of subspaces that might not be subspaces, but this is harder to picture). – Muphrid Aug 01 '14 at 14:36