I am having significant difficulty understanding where tensors fit into the big picture. I would really like to see exactly how tensors relate to simpler concepts that I am familiar with such as vector spaces, matrices and operations such as the inner product. Below I will outline to the best of my understanding what a tensor is and note any places where I am hitting a wall. And any all help on any particular section would be appreciated, I understand this is a long one.

Vector and dual spaces

I understand a vector space as a set of objects called vectors paired with a field on which the operations of vector addition and scalar multiplication can be defined and satisfy a set of axioms.

Every vector space has a corresponding "dual space" in which "covectors" live and these covectors can be seen as the set of all linear functions that map the vectors in the vector space onto the field over which both spaces are defined (in this case the real numbers):

$$v\in V, v^*\in V^*$$

$$v^*(v)=a , a\in \Bbb R$$

Question 1: Does this relationship work both ways? If both vectors and covectors are considered (1,0)- and (0,1)-tensors respectively then are they not both linear maps from eachother onto the underlying scalar field? When considered as arrays they appear to produce different objects, scalars or matrices depending on which order they are in:

$$\begin{pmatrix}x^1 \\ x^2 \\ x^3\end{pmatrix}\begin{pmatrix}x_1 & x_2 & x_3\end{pmatrix}= \begin{pmatrix}x^1x_1 & x^1x_2 & x^1x_3\\ x^2x_1 & x^2x_2 & x^2x_3 \\ x^3x_1 & x^3x_2 & x^3x_3\end{pmatrix} \in M_{2,2}$$


$$\begin{pmatrix}x_1 & x_2 & x_3\end{pmatrix}\begin{pmatrix}x^1 \\ x^2 \\ x^3\end{pmatrix}= x_1x^1 + x_2x^2 + x_3x^3 \in \Bbb R$$

The first equation looks like the tensor product of a vector and a covector however the second just looks like the inner product, in what way can I represent a vector as a function that takes a covector as an argument and returns a scalar? Since this is how I have seen tensors defined in the past.

Combining vector and dual spaces with the tensor product

By combining vector spaces and dual vector spaces with the tensor product we can construct new vector spaces that we call tensor spaces. A (r,s)-tensor is then defined as the number of vector and dual spaces required to construct it:

$$T^{i_1,i_2...i_r}_{j_1,j_2...j_s}\in V\otimes V\otimes V ... \otimes V^* \otimes V^* \otimes V^* ...$$

where V is some vector space and V* is its dual space, in abstract index notation the tensor $T^r_s$ has $r$ contravariant indices as superscripts and $s$ covariant indices as subscripts. The tensor $T$ is considered an element of the tensor space as defined above in the same way the vector $v$ is considered an element of the vector (or more generally tensor) space $V$.

If we have a tensor $T^{\mu}_{\alpha\beta}$ this is an element of the tensor space constructed as:

$T^{\mu}_{\alpha\beta}\in V\otimes V\otimes V^*\otimes V^*$

This is a (1,2) rank-3 tensor that could be represented as a 3-dimensional matrix.

The tensor, like a vector, also has an associated linearly independent basis set. For a vector the basis transforms covariantly, for covectors the basis transforms contravariantly and for this tensor $T^{\mu}_{\alpha\beta}$ its basis is constructed from the basis of the underlying vector and dual spaces:

$$T=t^{\mu}_{\alpha\beta}e_\mu\otimes e^\alpha\otimes e^\beta$$

Where $t^{\mu}_{\alpha\beta}$ are the components of the tensor and the basis set $e_\mu\otimes e^\alpha\otimes e^\beta$ can be through of as a set of 3-dimensional matrices that form a basis for all tensors in the tensor space above.

Question 2: Is this an acceptable definition of a tensor? And if so, is there any mathematical difference between a vector space and the tensor space constructed from it (and its dual) other than the dimensionality of the basis vectors?

The metric tensor

I see the metric tensor often highlighted as a special (0,2)-tensor, however I am tempted to consider it "just another (0,2)-tensor" defined on some vector space V and its dual V*. I understand that the metric is used to define angles and lengths in vector spaces, and given that it is a (0,2)-tensor it can be thought of as a bilinear map between two vectors and scalar, which, to my understanding is synonymous with the inner product in this case.

We can define a special type of vector space, an inner product space, by granting the vector space a third operation, the inner product.

Question 3: Could I define the metric tensor as the following:

Firstly I construct the set of (0,2)-tensors from the dual space in the same way I have defined T above:

$$G=V^*\otimes V^*$$

I then take from this tensor space one specific tensor which, in a given basis, defines angles and lengths between vectors in the vector space on top of which it is constructed. Such that:

$$g_{\mu\nu}\in G$$


$$g_{\mu\nu}v^{\mu}v^{\nu}\in \Bbb F$$

I then perform the same construction for the set of (2,0)-tensors on V and V*:

$$G'=V\otimes V$$

which, as a (2,0)-tensor is a bilinear map that takes two covector arguments an outputs a scalar such that

$$g^{\mu\nu}v_{\mu}v_{\nu}\in \Bbb F$$

Am I right in thinking that the metric tensor is one specific (0,2)-tensor that handles the inner product operation or is "the metric tensor" the set of all (0,2)-tensors in $G$? Or $G'$ in the case of the inverse metric tensor.

Contraction and tensors acting on other tensors

This is something that I find all around confusing and as such do cannot really even take a stab at. I want to say that we can relate the different types of tensors constructed on a vector field and its dual by three operations: index lowering, index raising and index contraction. I see broadly how these operations behave but only really at a notation level by having seen equations such as:


Would I be wrong if I rewrote this equations as


and assume this to be a summation over the $j$ components of $A_{ij}^{j}$?

I appreciate anyone who took the time to read any of this and again any help on any section is appreciated.

  • $\begingroup$ Didn’t you encounter the moment of inertia tensor when you studied classical mechanics? All it does is provide a linear but in general non-parallel relationship between the angular momentum vector and the angular velocity vector. In my opinion, dual spaces, covectors, and all the other mathematical concepts surrounding tensors can obscure the basic physics. $\endgroup$
    – G. Smith
    Commented Dec 12, 2019 at 0:59
  • $\begingroup$ And then didn’t you encounter them again when you studied advanced electromagnetism and learned how the field tensor provides a linear-but-not-parallel relationship between the four-velocity and the four-acceleration of a charged particle? $\endgroup$
    – G. Smith
    Commented Dec 12, 2019 at 1:03
  • $\begingroup$ I'm in a slightly strange position in that my degree is in chemistry and I am in a sense "converting" in to physics. I am learning 3rd year "maths methods in physics" which is requiring me to learn most of the foundational material myself. I had never in my life encountered tensors until I had them introduced in the context of special relativity. $\endgroup$
    – Charlie
    Commented Dec 12, 2019 at 1:06
  • $\begingroup$ OK. That makes things more difficult. There is a reason why the standard physics curriculum has the order it does. :) $\endgroup$
    – G. Smith
    Commented Dec 12, 2019 at 1:10
  • $\begingroup$ Yes I am discovering this fairly brutal reality in the real time. I think chemistry > physics is a particularly difficult transition since most of the intuitions for things I've learned in chemistry are in some way "abstractions" of the underlying physics/maths which means I am not only learning new things but having to entirely reconsider how I think about most things. $\endgroup$
    – Charlie
    Commented Dec 12, 2019 at 1:14

2 Answers 2


Tesseract has already provided the fancy math, so I’ll avoid symbols entirely and talk about simple physics. In my opinion, dual spaces, covectors, and all the other mathematical concepts surrounding tensors can obscure the basic physics.

A first exposure to tensors is usually by encountering the moment of inertia tensor when studying classical mechanics. All it really does is provide a linear but in general non-parallel relationship between the angular momentum vector and the angular velocity vector of a rotating object.

The next encounter is usually when you study advanced electromagnetism and learn how the field tensor provides a linear-but-not-parallel relationship between the velocity four-vector and the acceleration four-vector of a charged particle.

Do you see a pattern?

Don’t conclude that tensors are just for linearly mapping vectors to vectors. Do conclude that they are about linearly mapping physical quantities onto other physical quantities.


1) Yes. The dual space to a tangent space $T_p$ (denoted $T^*_p$)can be defined as the set of all linear maps that take members of $T_p$ to $\mathbb{R}$, i.e. if we have $\omega\in T^*_p$, $V,W\in T_p$, and $a,b\in\mathbb{R}$ $$\omega(aV+bW)=a\omega(V)+b\omega(W)\in\mathbb{R}$$ In the same way, since $T_p$ is dual to $T^*_p$, we can say that vectors are linear maps mapping dual vectors to $\mathbb{R}$, where $V(\omega)\equiv\omega(V)=\omega_\mu V^\mu$.

2) Not exactly. You use a Cartesian product to define the map, but the tensor product in coordinate space. A $(k,l)$ tensor is defined as a multilinear map between a vector space/dual vector space as follows: $$T:T_p\times\cdots\times T_p\times T^*_p\times\cdots\times T^*_p\to\mathbb{R} $$ where there are $k$ $T_p$ and $l$ $T^*_p$. This becomes, in coordinate space ($\hat{e}_{(\mu)}$ is the basis for $T_p$, $\hat{\theta}^{(\mu)}$ is the basis for $T^*_p$): $$T=T^{\mu_1...\mu_k}_{\;\;\,\;\;\;\;\;\nu_1...\nu_l}\;\hat{e}_{(\mu_1)}\otimes\cdots\otimes \hat{e}_{(\mu_k)}\otimes\hat{\theta}^{(\nu_1)}\otimes\cdots\otimes\hat{\theta}^{(\nu_{l})}$$

3) You are correct; the metric tensor is one particular element of this space.

4) No, this is not correct. Note that this uses the Einstein summation convention: when there is the same index 'upstairs' and 'downstairs', we sum over this. Thus, the product of some tensors should only have the independent variables on the other side. So, $g_{\mu\nu}A^\mu$ would be $$\sum_\mu g_{\mu\nu}A^\mu=g_{0\nu}A^0+g_{1\nu}A^1+g_{2\nu}A^2+g_{3\nu}A^3$$ As you can see, this expression is a sums to a dual vector.* Because this is the metric tensor, this expression is exactly equivalent to $A_\nu$. However, this does not hold for arbitrary tensors, but the expansion can be carried out for any two arbitrary tensors. For example, $$\omega_\mu V^\mu = \omega_0 V^0 + \omega_1 V^1 + \omega_2 V^2 + \omega_3 V^3$$ $$T^{\mu\nu}_{\;\;\;\sigma}S^{\sigma\rho}_{\;\;\;\alpha}=T^{\mu\nu}_{\;\;\;0}S^{0\rho}_{\;\;\;\alpha}+T^{\mu\nu}_{\;\;\;1}S^{1\rho}_{\;\;\;\alpha}+T^{\mu\nu}_{\;\;\;2}S^{2\rho}_{\;\;\;\alpha}+T^{\mu\nu}_{\;\;\;3}S^{3\rho}_{\;\;\;\alpha}$$ As previously noted, the metric tensor has the special ability to raise and lower indices: $$g^{\mu\alpha}T_{\mu}=T^{\alpha}$$ $$g^{\nu\alpha}g_{\mu\beta}T^{\mu}_{\;\;\nu\rho}=T^{\;\alpha}_{\beta\;\,\rho}$$ One can finally contract tensors as well, an operation that takes $(k,l)$ tensors to $(k-1,l-1)$ tensors, if there are similar indices on the top and bottom: $$T^{\mu\nu}_{\;\;\;\nu\rho}=S^{\mu}_{\;\,\rho}$$ There are a couple other operations, but they don't have much to do with raising\lowering indices, so I won't mention them here.

*Note that acting tensors on each other doesn't always result in another tensor. It will for the metric tensor, however.

  • $\begingroup$ 1) So I would be correct in saying that the dual of V, V* is the set of all linear maps from V to $\Bbb R$ and that the dual of V* which is equivalent to V is again the set of all linear maps that take V* to $\Bbb R$? $\endgroup$
    – Charlie
    Commented Dec 12, 2019 at 1:08
  • $\begingroup$ 2) I see my mistakes here thanks for the correction. $\endgroup$
    – Charlie
    Commented Dec 12, 2019 at 1:09
  • $\begingroup$ 3) Just one other thing that bugs me, when we change basis the components of the metric tensor change predictably as is the case for all tensors, but is this still the same tensor underneath? Or is one metric tensor chosen per basis set. (Upon reading what I've written here It might not be clear what I'm actually asking, I might just have to think about this one, my bad). $\endgroup$
    – Charlie
    Commented Dec 12, 2019 at 1:17
  • 1
    $\begingroup$ 1) Absolutely! 3) A tensor is an abstract geometrical object. It's rather like how, if I change coordinate systems, a vector in $\mathbb{R}^3$ stays the same object, but has different numbers describing it. This is the same concept with the metric tensor; one tensor, different coordinate systems. 4) That was my mistake, and I will edit this. $\endgroup$ Commented Dec 12, 2019 at 1:55
  • 1
    $\begingroup$ No problem mate :). Do note that some users might scold you for thanking folks though, so take note of that. $\endgroup$ Commented Dec 12, 2019 at 2:03

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