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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 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 intepreted as a vector or (0,2) as a bivector and in general (0,s) tensor as n-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 ? Please dont close this post. My answer is not in "What is tensor ?"

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Just like $(r,0)$ are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors a scalar, and you seem to accept this visualization, the $(r,s)$ tensors are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors and $s$ vectors i.e. $(1,0)$ tensors a scalar.

This is just one way among many how to present or visualize given tensors. It is a way that suggests that one may "contract" all the tensor's indices with vectors to get a scalar, i.e. a $(0,0)$ tensor, as a result. But there are many other ways how the indices of tensors may be contracted to get other tensors (including scalars, vectors, covectors, and other tensors). You presented one such interpretation of the $(1,1)$ tensors – they're operators, either on the space of vectors or co-vectors.

Quite generally, the most general product of $(r_i,s_i)$ tensors for $i=1,2,\dots N$ without any contraction of indices is a tensor of the type $$ (\sum_{i=1}^N r_i, \sum_{i=1}^N s_i) $$ which in plain English means that the number of lower and upper (covariant and contravariant indices) is simply being added from the individual products.

Concerning the elementary non-trivial (non-scalar) tensors, $(1,0)$ tensors and $(0,1)$ tensors are vector spaces that are "dual" (a couple of a sort) to each other, in the sense of linear algebra. One of these spaces contains all the linear forms acting on the other space, and vice versa. The relationship is symmetric, therefore it is a "duality". So if you think that you know an interpretation of one of these tensors, you should admit that you know an interpretation of the other space (and its elements, tensors), too.

Once you have "interpretations" for the elementary $(1,0)$ and $(0,1)$ tensors, you may have an "interpretation" for the most general $(r,s)$ one as a "multilinear functional" acting on $r$ vectors of one kind and $s$ vectors of the dual type.

Luboš Motl
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    Thank you for your attention. I think my question is still unresolved. To clarify your answer to me, first please explain to me how (0,1) tensor interpreted as a vector; that is why a linear function of a functional interpretd as a vector ? –  Jul 31 '14 at 13:57
  • The dual to the dual of a vector is (isomorphic to) the same vector. Mathematicians love to say lots of fancy things about it, see e.g. http://math.stackexchange.com/questions/540020/dual-to-the-dual-norm-is-the-original-norm - but it is really a trivial fact. – Luboš Motl Jul 31 '14 at 15:49
  • Ok. I agree that dual of a dual of a vector space is isomorphic to the vector space, but this doesnt mean dual of the dual of a vector is exactly the same vector, how you can show that a vector can be expressed as a function of a linear functional ? –  Jul 31 '14 at 15:57
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    Sorry, there is nothing such as a "dual of an individual vector". That's really the point. If you could map the vectors one by one, then there would be no reason to consider the spaces to be different. Now, if you have an inner product on the space, then $(1,0)$ and $(0,1)$ vectors are really the same - there is a map for vectors on one-to-one basis - but then the distinction between $(r,s)$ and $r+s$ indices becomes silly. The $(r,s)$ are only talked about if no canonical important universal inner product exists, and then there is no way to "dualize" individual vectors! – Luboš Motl Jul 31 '14 at 16:01
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    The isomorphism only applies to the spaces. If I have a "linear form on the space of linear forms on the space V", I may identify it with an element of "V" itself. But V and dual V - the space of (0,1) and (1,0) vectors - are really not isomorphic to each other so it's totally wrong to identify individual vectors in them. – Luboš Motl Jul 31 '14 at 16:02
  • Thank you. But i think my problem still remains. I understood (1,0) tensor interpreted as covector, since (1,0) tensor definition and covector definition coincided. But how about a vector and (0,1) tensor. I cant understand how these objects coincided. –  Aug 01 '14 at 07:42
  • Hi, if $(1,0)$ tensor is the covector, then the $(0,1)$ tensor is the dual one, the normal vector or contra-vector or whatever you want to call it. – Luboš Motl Aug 01 '14 at 08:49
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    However, I cant understand how a vector can be viwed as a linear function of a functional. –  Aug 01 '14 at 09:17
  • You wrote "I agree that dual of a dual of a vector space is isomorphic to the vector space" and it's exactly the same statement. So you should decide whether you understand it or not. – Luboš Motl Aug 01 '14 at 09:56
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    Isomorphism between two space means the structure of two spaces without confining ourseleves to their elements is the same. This doesnt mean their elements are the same. For example any two n-dimensional vector space are isomorphics. Does this mean that their elements are the same ?! –  Aug 01 '14 at 10:07
  • Yes, they're the same. There exists a canonical isomorphism. I agree with you that there may exist non-unique isomorphisms between the two structures, i.e. the isomorphism is undetermined up to a (large) group of automorphisms, but the isomorphism of a space and the dual of its dual is absolutely canonical. Yes, the elements are the same. The duality is nothing else than moving the index to the other place - from bottom to top or vice versa. – Luboš Motl Aug 01 '14 at 11:49
  • It's really trivial to show you what the form on the space of forms is. Take a vector V on the original space. Then you have forms F on the dual space that give you a real number F(V). Now, I want a form G on the space of forms F, i.e. G(F). For every form F, I want a real number G(F), OK? It's easy to define a "form on the space of forms" that is uniquely associated with a vector V on the original space, namely by G_V(F)=F(V). ;-) For each V, I can define such a G_F that assigns a number to any form F. – Luboš Motl Aug 01 '14 at 11:52
  • Let's take the original space "real numbers". Do you agree the elements of dual of the dual of our vector space are "functions". And "numbers" and "functions" are two different things. Am I right ? –  Aug 01 '14 at 12:23
  • No, by the dual space, we mean the space of linear functions, and they are given by one parameter. To make your example more insightful, consider the original vector space to be the space of $x$ coordinates in meters. So "x=5 meters" and "x=-3 meters" are elements. The dual space contains linear functions $f: x\to R$ which obviously have the form $f=ax$. The result is real, unitless, so $a$ must have units of inverse meters by dimensional analysis, OK? That's why you can't identify the elements of the original space with those in the dual space. – Luboš Motl Aug 01 '14 at 13:34
  • But both of these spaces are one-dimensional spaces. The dual to dual contains linear functions that map $a$, with the units of inverse meters, to real numbers. It's clearly just the multiplication of $a$ by another coefficient whose unit is again 1 meter. Those may be canonically identified with the original elements x. Please try to think more than inventing error-rich excuses. This is my last answer here because the time needed to settle this complete triviality has become excessive. – Luboš Motl Aug 01 '14 at 13:36