Jtdylktuy

Taking into account what has been said here, recall that if ${e_1,...,e_n}$ is a basis of $V$ and ${omega^1,...,omega^n}$ its dual basis, then any $Tinmathcal T^{(p,q)}$ can be written:

begin{equation} T=sum lambda_{i_1, cdots, i_p}^{j_1, cdots, j_q} e_{i_1} otimes cdots otimes e_{i_p} otimes omega^{j_1} otimes cdots otimes omega^{j_q} end{equation}

Think about the simple cases. Tensors of type $(0,1)$ are linear maps $T:Vrightarrow K$, that is, elements of $V^*$. Now, tensors of type $(0,2)$ are usually called bilinear forms. They are multilinear maps $T:Vtimes V rightarrow K$. Now, the easiest way of building a $(0,2)$-tensor is picking two covectors $f_1$ and $f_2$, and use tensorial product:
$$T=f_1otimes f_2$$
(note that I wrote sub-indices, there is no difference between writing super-indices, and it's advisable to do it in the general case, but for $mathcal T^{(0,2)}(V)$ it's all right).

Now, using the first equation, every $(0,2)$-tensor is like this:
$$T=sum lambda_{ij}  omega^iotimes omega^j$$
(recall that ${omega^1,...,omega^n})$ is a basis of $V^*$.
So now you can see that $mathcal T^{(0,2)}(V)=V^*otimes V^*$.

In the general case, all tensors are sums of multiples of tensors like this:
$$e_{i_1} otimes cdots otimes e_{i_p} otimes omega^{j_1} otimes cdots otimes omega^{j_q} qquad qquad (1)$$

so this is why $mathcal T^{(p,q)}(V)=Votimes overset{p text{ times}}{...} otimes V otimes V^*otimes overset{q text{ times}}{...} otimes V^*$

Let us focus on tensors like $(1)$.
If $v_1, ..., v_qin V$ and $xi^1, ... xi^pin V^*$, then

$$e_{i_1} otimes cdots otimes e_{i_p} otimes omega^{j_1} otimes cdots otimes omega^{j_q}(xi_1, ... xi_q,v_1, ..., v_p,)=e_{i_1}(xi^1)cdots e_{i_p}(xi^p) cdot omega^{j_1}(v_1)cdots omega^{j_q}(v_q)$$

as I suppose you know. So, in a certain way, you could say that these $omega$'s are waiting for a vector to produce numbers that later will be multiplied.

Now, forget that you know the elements of $V^*$ are functionals. They now are, simply, vectors, (as they are elements of a vector space), and its dual is $V^{**}=V$. So the $e$'s, considered as functionals over $V^*$ are also waiting for a vector (but now a vector is an element of $V^*$) to produce numbers that will be multiplied.

The difficulties with this issues is that one has to keep in mind the different points of view. It is true that $V^*$ is the dual of $V$, and the elements of $V^*$ are functionals over $V$. But, as $V^*$ is again a vector space, we can forget for a moment about $V$, considering the elements of $V^*$ as simple vectors, and take the dual $V^{**}$. So, in the end, vectors and covectors behave in really similar ways.

Regarding the isomorphism $phi: Vto V^{**}$, if we pick $vin V$. So now, $phi(v)$ is a functional in $V^*$, and to define who $v$ is, we have to say what is the value of $phi(v)(omega)in K$ for every $omegain V^*$. But we know $omega$ is a functional on $V$, so $omega(v)in K$. So we use this fact to define $phi(v)$:

$$phi(v)(omega)=omega(v)$$

In the end, we would end writing $phi(v)=v$ so in the end, we get $$v(omega)=omega(v)$$
and this is why the theorem that states the isomorphism $Vto V^{**}$ is called reflexivity theorem. When you see $v(omega)$, just think it is the same as $omega(v)$.

The space of linear maps $f:Vto k$ is $V^*$. As such, we can show that the space of multilinear maps from $Pi_i^n M_i$ to $k$ will be exactly $M_1^*otimes M_2^*otimesdots otimes M_n^*$, which is exactly what is going on here.

So the space of maps from $Pi_1^p Vtimes Pi_1^q V^*to k$ would be $bigotimes_1^p V^*otimes bigotimes_1^q V^{**}$

Since this definition seems to implicitly be working with finite dimensional space $V^{**}$ can be replaced with $V$.

I will probably not add much new, just use a slightly different language.
A $V^*$-terms in equation $(2)$ for $(p,q)=(0,1)$ clearly represents a linear maps $Vto K$ (in eq $(3)$)). Similarly, $V$ (in equation $(2)$ with $(p,q)=(1,0)$) can be canonically identified with $V^{**}$, elements of which are by definition linear maps $V^*to K$ (that is $(3)$ for $(p,q)=(1,0)$).

Using the notation common in physics, you can simply assume a fixed basis and represent vectors in coordinates like $v^i:=(v^1,ldots, v^n)$ and duals like $alpha_i:=(alpha_1,ldots,alpha_n)$ and their pairing $alpha_i v^i:=sum_i alpha_i v^iin K$. Then you can ask "what is $v^i$ (a vector!) doing?" Answer: it takes $alpha_i$ and returns $alpha_i v^i$. Similarly, what is $alpha_i$ (a dual!) doing? Answer: it takes $v^i$ and returns $alpha_i v^i$. This construction of course directly generalizes. What is $T^{i_1ldots, i_p}_{j_1ldots j_q}$ doing? It takes $(v_1)^{j_1},ldots, (alpha^p)_{i_p}$ --- in other words, $q$ vectors and $p$ covectors --- and returns the corresponding sum
$$
sum_{i_u,j_v} T_{j_1ldots}^{i_1ldots} ,,(v_1)^{j_1}ldots (alpha^p)_{i_p}
$$
The reason is that you always want to pair upper- (vector-) indices with lower indices and vice versa.

Further, you ask

Can we say then that the $q$ elements $V^*otimes V^*otimescdots$ in Eq. (2) are linear functionals waiting for the same number of
vectors $V$ to produce real numbers that are later multiplied?

Here you have to be more careful. For convenience, I will restrict myself to $V^*otimes V^*$ as it immediately generalizes. In some sense, you are right and the object $T_{ij}$ (an element of $V^*otimes V^*$) is "waiting" for two vectors $v,w$ on which it will be evaluated to $sum T_{ij} v^i w^j$.

However, if you have an element of $V^*otimes V^*$, there is no way to identify "two elements" of $V^*$ that it "represents". Note that you have a natural map $V^*times V^*to V^*otimes V^*$ that to two elements $(alpha,beta)$ assigns $alphaotimesbeta$ which acts on vectors via
$$
(alphaotimesbeta)(v,w):=alpha(v) beta(w).
$$
But

1. This map is not surjective. For example, a scalar product on $Bbb R^2$ cannot be represented this way (exercise 1 for you). In fact, a tensor is of this form iff the matrix $T_{ij}$ defined above in coordinates, has rank 1 (exercise 2 for you)


2. This map is even not linear! (assuming the linear structure on $V^*times V^*$ such as $(alpha,beta)+(alpha',beta')=(alpha+alpha', beta+beta')$). Try to verify that $(alphaotimes beta)(v,w)+(alpha'otimesbeta')(v,w)neq (alpha+alpha')otimes (beta+beta')(v,w)$ in general (exercise 3 for you).


3. This map, however, is multilinear: $(alpha+alpha')otimes beta=alphaotimes beta+alpha'otimes beta$ and similarly in the second component (exercise 4)


4. There is another, maybe more natural, definition of tensor product that you may like more and that completely avoids the "duals". You can consider $V^*otimes V^*$ to be an abstract vector space freely generated by vectors $alphaotimes beta$ where $alpha,betain V^*$ where you have only the relations $(alpha+alpha')otimes beta=alphaotimes beta + alpha'otimes beta$, $(kalpha)otimes beta=k(alphaotimes beta)$ etc. In other words, you can just get used to these relations, completely ignoring a meaning of these objects. Then you don't need all those duals that bother you. (Exercise 5: show the equivalence of these two approaches)

Further, I have two general recommendations. Think of a bilinear form, which is a tensor of type $(0,2)$, that is, an element of $V^*otimes V^*$. In coordinates, just a "matrix" $T_{ij}$ that acts on two vectors. Maybe it helps to go around all these texts with this in mind. Maybe try to show (exercise 6) that a bilinear form can be identified with a linear map $Votimes Vto K$ (this is confusing, I know)

The second recommendation: as this material is very standard and taught at all universities, it is questionable whether the best way to learn it is just from the internet (if you do so, I don't know). Maybe going over a book or some Coursarea course might help.

What are all the $(1,0)$ tensors on $V$? They're linear functions from $V^{*}$ to $Bbb R$. It turns out that every such linear function is just "evaluate on an element of $V$". So they correspond, 1-to-1, with elements of $V$. (This doesn't work in infinite dimensions, but for finite dimensions--definitely.)

What about $(2,0)$ tensors on $V$? They're bilinear functions on $V^{*}$. So they take in two covectors and produce a number. One way to produce such a thing is to pick a pair of vectors $v$ and $w$ and write down

$$
T_{v,w}(phi, theta) = (phi(v)) cdot (theta(w))
$$

Of course, if you picked the pair $(2v, w/2)$, you'd get the same bilinear function. And also the same bilinear function for $(v/3, 3w)$, etc. In fact, EVERY bilinear function on $V^{*}$ looks like "evaluate on two vectors and take the product of the results" (or a sum of such things).

So these bilinear functions correspond exactly to elements of $V otimes V$.

Are you seeing the pattern here?

[This will never be the accepted answer, so no worries.]

I found in Introduction to Vectors and Tensors: Linear and Multilinear Algebra by Ray M. Bowen and C.-C. Wang the following exposition of the functions in the OP to be very enlightening, and I'm writing it here with minimal modifications:

Step 1:

If $mathbf{v}in mathscr V$ and $mathbf{v^*}in mathscr V^*$, their scalar product can be computed in scalar form as follows: Choose a basis ${mathbf{e}_i}$ and its dual basis ${mathbf{e}^i}$ for $mathscr V$ and $mathscr V^*$, respectively so we can express $mathbf{v}$ and $mathbf{v}^*$ in component form. Then we have

$$left<mathbf v^*, mathbf vright>=left<v_i;mathbf e^i, v^j ; mathbf e_jright>=v_i,v^j left<mathbf e^i, mathbf e_jright>= v_i,v^j; delta_j^i=v_i,v^i$$

Step 2:

If $mathscr V_1,cdots,mathscr V_s$ is a collection of vector spaces, then a $s$-linear function (multilinear function) is a function

$$mathbf A: mathscr V_1,cdots,mathscr V_srightarrow mathbb R$$

that is linear in each of its variables while the other variables are held constant. If the vector spaces $mathscr V_1,cdots,mathscr V_s$ are the vector space $mathscr V$ OR its dual space $mathscr V^*,$ then $mathbf A$ is called a TENSOR.

More specifically, a tensor of order $(p,q)$ on $mathscr V$, where $p$ and $q$ are positive integers, is a linear function

$$bbox[5px,border:2px solid aqua]{T^p_q,V equiv underset{ptext{ times}}{underbrace{mathscr V^*timescdotstimes mathscr V^*}}times underset{qtext{ times}}{underbrace{mathscr Vtimescdotstimes mathscr V}} rightarrow mathbb R}$$

Step 3:

If $mathbf v$ is a vector in $mathscr V$ and $mathbf v^*$ is a covector in $mathscr V^*$, then we define the function

$$mathbf v ,otimes, mathbf v^*: mathscr V^* times mathscr V rightarrow mathbb R$$

by

$$mathbf v ,otimes, mathbf v^*,(mathbf u^*,,,mathbf u)equiv left<mathbf u^*,mathbf vright>left<mathbf v^*,mathbf uright>$$

for all $mathbf u^*in mathscr V^*, mathbf uin mathscr V.$ Clearly $mathbf v ,otimes, mathbf v^*$ is a bilinear function, so $mathbf v ,otimes, mathbf v^*in T^1_1(mathscr V).$

Step 4:

The tensor product can be defined for arbitrary number of vectors and covectors. Let $mathbf v_1,cdots,mathbf v_p$ be vectors in $mathscr V$ and $mathbf v^1, cdots, mathbf v^q$ be covectors in $mathscr V^*.$ Then we define the function

$$mathbf v_1,otimescdotsotimes,mathbf v_p,otimesmathbf v^1cdotsotimes,mathbf v^q:mathscr V^*,times,cdots,timesmathscr V^*,timesmathscr V,timescdots,times,mathscr Vrightarrow mathbb R$$

by

$$bbox[5px,border:2px solid aqua]{begin{align}&mathbf v_1,otimescdotsotimes,mathbf v_p,otimesmathbf v^1cdotsotimes,mathbf v^q,left(mathbf u^1,cdots,mathbf u^p,mathbf u_1,cdots,mathbf u_qright)\[2ex]
&equivleft<mathbf u^1,mathbf v_1 right>,cdots left<mathbf u^p,mathbf v_p right>,left<mathbf v^1,mathbf u_1 right>,cdotsleft<mathbf v^q,mathbf u_q right>,
end{align}}$$

for all $mathbf u_1,cdots,mathbf u_qin mathscr V$ and $mathbf u^1,cdots,mathbf u^qin mathscr V^*.$ Clearly this function is $(p+q)$-linear, so that

$$mathbf v_1 otimes cdotsotimesmathbf v_potimesmathbf V^1otimescdotsotimes mathbf v^q in T^p_q(mathscr V)$$

is called the TENSOR PRODUCT of $mathbf v_1,cdots,mathbf v_p$ and $mathbf v^1,cdots,mathbf v^q.$