# How to reconcile these two tensor notations?

How can I connect the symbols, i.e. the notation, preferably in English or with a 2 x 2 (or 3 x 3) matrix example, between an order 2 tensor expressed as:

A $$(p,q)$$ tensor, $$T$$ is a MULTILINEAR MAP that takes $$p$$ copies of $$V^*$$ and $$q$$ copies of $$V$$ and maps multilinearly (linear in each entry) to $$k:$$

$$T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times \cdots \times V\times V}} \overset{\sim}\rightarrow K\tag 1$$

and

$$\large \mathbf{T}= T_{ij}\;\mathbf{\hat e_i}\otimes\mathbf{\hat e_j}\tag 2$$

?

Would $$\large T_{ij}$$ in Eq.2, which I guess can be interpreted as coefficients or entries in a matrix) be the $$V^*$$ elements of the dual space (functionals), while the $$\mathbf{\hat e_i}$$ and $$\mathbf{\hat e_j}$$ are the vectors $$V$$? Are the $$\times$$ in Eq.1 Cartesian products (presumably they can't be cross-products...)? Are the $$V$$'s in Eq. 1 just vectors, or are they elements of the double dual? Are the indices $$(p,q)$$ in Eq.1 the equivalent of $$(i,j)$$ in Eq.2?

I realize Eq. 1 is probably more general, but it should be possible to reduce it to the more simple case of Eq.2, again just to be able to enunciate what the symbols are. Do both equations produce a field element?

• I realize that I am asking about the same issue from different perspectives, and that I haven't accepted any answers so far. It is extremely rare for me not to "accept", but I am hoping to get an answer that at least anchors the problem, and charts the road ahead in the understanding of this issue. After that, I will go back and accept the previous answers. Feb 12, 2017 at 17:35
• In case it's helpful, the first time I understood tensors (including the issues that seem to be puzzling you) was while working through Chapter 4 of Spivak's Calculus on Manifolds. Feb 12, 2017 at 17:55
• @AndrewD.Hwang I have posted 3 questions on this topic over the weekend. Your answer and follow-up have been great, so I would like to bring to your attention a bounty of hard-earned 100 point I just posted on my original question on this topic. Feb 12, 2017 at 20:21
• see a version math.stackexchange.com/questions/1545870/… Feb 14, 2017 at 21:22
• Or this another math.stackexchange.com/questions/1750015/… Feb 14, 2017 at 22:37

The $\times$ in eq. 1 are Cartesian products. Note that, in finite dimensions, $V^{**}=V$, so vectors can be seen as elements of the double dual.

Now, if $\mathcal T^{(p,q)}(V)$ denotes the space of $(p,q)$ tensors over a vector space $V$, $\mathcal T^{(p,q)}(V)$ is a vector space, and if we pick $\{e_1, \cdots, e_n\}$ a basis for $V$, and $\{\omega_1, \cdots, \omega_n\}$ the dual basis, we can construct a basis for $\mathcal T^{(p,q)}(V)$ with the elements:

$$e_{i_1} \otimes \cdots \otimes e_{i_p} \otimes \omega^{j_1} \otimes \cdots \otimes \omega^{j_q}$$

with $\{i_1, ..., i_p, j_1, ..., j_q\}\in\{1,...,n\}$.

So if $T\in \mathcal T^{(p,q)}(V)$, then it would be $$T=\sum \lambda_{i_1, \cdots, i_p}^{j_1, \cdots, j_q} e_{i_1} \otimes \cdots \otimes e_{i_p} \otimes \omega^{j_1} \otimes \cdots \otimes \omega^{j_q}$$

Equation 2 is the particular case of the previous equation for $\mathcal T^{(0,2)}(V)$ (we will write, instead of $i_1,i_2$, $i,j$):

$$T=\lambda_{ij}\ \omega^i \otimes \omega^j$$

As you only have two indices, you can form a matrix with the numbers $(\lambda_{ij})$.

• What is the $0$ in $\mathcal T^{(2,0)}(V)$? And the $w$'s in the final expression have supraindices, as opposed to subindices in Eq.2... Feb 12, 2017 at 17:50
• That means that your tensor is $T:V\times V \longrightarrow K$, it is, you won't take any copies of $V^*$. With your notation, it would be $\mathcal T^{(0,2)}$. Let me fix it. Feb 12, 2017 at 17:51
• Why is your last expression different from Eq.2 (sub- vs. supra- indices)? Feb 12, 2017 at 17:53
• It doesn't matter where you put the indices. If you are working with a $(p,q)$ tensor, it's common to write them as I've done: sub for $e$'s, and supra for $\omega$'s. But it's just for the sake of clarity. $$T=\lambda_{ij} \omega_i \otimes \omega_j$$ is perfect. Feb 12, 2017 at 17:56
• Yes, both notations mean the same. No matter where you write the indices. Feb 12, 2017 at 18:05