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I am currently watching this excellent video series building up to general relativity. We have finally started looking at tensors and a question came up from the audience which (to my understanding) was asking why tensors are defined as multi-linear maps from sets of vectors and covectors to the real numbers yet (1,1)-tensors are often viewed as linear maps from vectors to vectors.

He says that the two ways of viewing the (1,1)-tensor contain the same information, but I don't understand how. If anyone could provide another explanation for why this is true I would really appreciate it.

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This kind of trick is common when thinking about tensors.

Let $v$ be a vector, $\alpha$ a one-form, and $c$ a number. A vector gives a map from one-forms to numbers, $$\alpha \mapsto v(\alpha)$$ but it also gives a map from numbers to vectors, $$c \mapsto c v.$$ Similarly, a $(1, 1)$ tensor $T$ gives a map from a vector and covector to numbers, $$(\alpha, v) \mapsto T(\alpha, v).$$ But it also gives a map from vectors to vectors, $$v \mapsto T(\cdot, v).$$ To see that the output here is a vector, note that it gives a map from one-forms to numbers, $$\alpha \mapsto T(\alpha, v).$$ By continuing with this, you can also use a $(1, 1)$ tensor to define a map from covectors to covectors, or numbers to a vector and covector, and so on.

To avoid having to associate the exact same object with a ton of different maps, physicists use index notation. In this notation, the five preceding maps take the form $$\alpha_i \mapsto v^i \alpha_i, \quad c \mapsto c v^i, \quad (\alpha_i, v^j) \mapsto T^i_j \alpha_i v^j, \quad v^i \mapsto T^i_j v^j, \quad \alpha_i \mapsto (T^i_j v^j) \alpha_i.$$ From this you can get the general pattern, which is that a $(n, m)$ tensor can act on a $(n', m')$ tensor to yield a $(n+n'-k, m+m'-k)$ tensor where $k$ is the number of upstairs/downstairs index pairs that are contracted.

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  • $\begingroup$ If a tensor is capable of providing so many different maps why are tensors almost always defined as objects that take vectors and covectors and produce scalars? Also where you've written $$v \mapsto T(\cdot, v).$$ I don't see what this is doing, this seems like a very strange operation. Which tensor space does $T(\cdot,v)$ belong to relative to the original vector space V? $\endgroup$ – Charlie Dec 25 '19 at 23:32
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    $\begingroup$ @Charlie As I show in the next line, $T(\cdot, v)$ is a vector, because it is a linear map that takes a covector and returns a number. $\endgroup$ – knzhou Dec 25 '19 at 23:46
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    $\begingroup$ @Charlie That definition is the cleanest starting point, and it shields people from having to think about all the other closely related maps while the foundation is laid. But in practice you end up using every single one of the maps I mentioned, with about equal frequency. $\endgroup$ – knzhou Dec 25 '19 at 23:47
  • $\begingroup$ What's happening in $ \alpha_i \mapsto (T^i_j v^j) \alpha_i$? Where does the $ v^j $ come from? I would have expected $ \alpha_i \mapsto T^i_j \alpha_i$. If that's not a mistake, could you please elaborate on its meaning a bit? $\endgroup$ – Christian Aichinger Dec 26 '19 at 8:37
  • $\begingroup$ @ChristianAichinger This is an example of how $T(, \cdot v)$, i.e. $T$ with one argument already plugged in, is a vector. $\endgroup$ – knzhou Dec 26 '19 at 17:31
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A $(n,m)$ tensor eats $n$ vectors and $m$ co-vectors and spits out a real number. A $(1,1)$ tensor eats one vector and one co-vector and spits out a real number. Let's call our $(1,1)$ tensor $f(.,.)\colon V \times V^{*} \mapsto \mathbb{R} $. The first argument is a vector and the second argument a co-vector. So for generic vectors $a \in V, b \in V^{*}$, $f(a,b) \in \mathbb{R}$.

What about $f(a,.)$? Clearly, it eats a co-vector (in the empty argument) and spits out a real number. But this is precisely the definition of a vector. So, $f(a,.)$ is a vector. Therefore $f(.,.)$ ate a vector(i.e $a$) and produced a vector $f(a,.)$.

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  • $\begingroup$ Would I be right in saying that in this case $f(a,.)$ is a member of the double dual space $V^{**}$ which is isomorphic to V? $\endgroup$ – Charlie Dec 25 '19 at 23:26
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    $\begingroup$ Yes, that is correct. $\endgroup$ – Anonjohn Dec 25 '19 at 23:35

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