# Definition of tensors

I'm studying the book "Riemannian Geometry" by Petersen and since I'm new to the subject, I'm helping myself also with the more introductory DoCarmos's book. I'm a bit confused about the definition of tensor. The definition given by DoCarmo is the following.

Given a Riemannian manifold M, a tensor $T$ of order $r$ is a multilinear map $$T: \underbrace{\mathcal{X}(M) \times \dots \times \mathcal{X}(M)}_{r \text{ times}} \to \mathcal{D}(M)$$ where $\mathcal{X}(M)$ is the set of all the smooth vector fields over $M$ and $\mathcal{D}(M)$ is the set of all the smooth functions on $M$.

But Petersen's book talks about $(q, r)$-tensors and I guessed from the context that that for him a $(q, r)$-tensor is a multilinear map

$$T: \underbrace{\mathcal{X}(M) \times \dots \times \mathcal{X}(M)}_{r \text{ times}} \to \underbrace{\mathcal{X}(M) \times \dots \times \mathcal{X}(M)}_{q \text{ times}}$$.

Then I took a look also to John Lee's book and I found this definition.

A $r$-covariant and $q$-contravariant tensor on a real vector space $V$ is a multilinear map $$T: \underbrace{V^* \times \dots \times V^*}_{q \text{times}} \times \underbrace{V \times \dots \times V} \to \mathbb{R}.$$

Then he extend the definition to manifolds considering smooth sections of tensor bundles.

What is the differences between these approaches? Could someone explain me better how Petersen sees tensors?

• The definition of Petersen is at least unconventional. the standard definition i know is the one also given by Lee, which is equivalent the the one by DoCarmo if we restrict ourself to covariant quantities. – tired Feb 22 '16 at 13:45
• I'm not really sure if I have understood correctly what Petersen means, because he didn't give any definition. But for example he says that the curvature $R(X,Y)Z$ is a $(1,3)$-tensor, while the metric $g$ is a $(0, 2)$-tensor. – Onil90 Feb 22 '16 at 13:52

Given a $q$-contravariant $r$-covariant tensor in the sense of John Lee's book, You can get a different multilinear map from $q-1$ copies of $V^*$ and $r$ copies of $V$ to $V$ as follows.

Let $\bar{T}(\omega_2,\ldots,\omega_q,v_1,\ldots,v_r)$ be an element $w\in V$ such that $\omega_1(w) = T(\omega_1,\ldots,\omega_q,v_1,\ldots,v_r)$ for any $\omega_1$. Repeating this, we can get a map from $q-2$ copies of $V^*$ and $r$ copies of $V$ to $2$ copies of $V$, and so on until we get a multilinear map from $r$ copies of $V$ to $q$ copies of $V$. This gives a (natural) isomorphism between $q$-contravariant $r$-covariant tensors to multilinear maps from $q$ copies of $V$ to $r$ copies of $V$.

That's the "tricky" part. Extending that from vectors and scalars to vector fields and functions gets you to Petersen's definition. That's a little hand-wavy, but hopefully puts you on the right track.

Edit: By the way, just picked up my copy of Lee's Riemmannian Manifolds book ( great book, second edition coming in 2017 according to rumour, can't wait ), and what I wrote above is basically Lemma 2.1

• Thank you! But is it better Petersen's definition or Lee's definition?Which is the most used one? – Onil90 Feb 22 '16 at 13:59
• i mean sure i can construct this kind of things, but personally i am not convinced this is a very useful way to actually define a tensor. – tired Feb 22 '16 at 14:22
• Hmm, I think it's a question of "flavor". It would be nice when you're learning if there were two different names for the two types of objects, but trust me, after using them for a couple of years, you'll be happier to refer to them by the same name. – Callus - Reinstate Monica Feb 22 '16 at 14:23
• @tired Which part don't you like? that it's not a map to the reals, or that it's defined using vector fields and functions. I think there are lots of scenarios where both are more natural, it just depends what kind of math you do, mainly. I spent a lot of time thinking about algebro-geometric things and now find Petersen's definition the most natural. Of course, if we go that direction, there's no reason to stop at just vector fields and functions, let's do tensor products of sheaves of $\mathcal{O}_X$-modules! It really depends on the "category" of stuff you're looking at I think. – Callus - Reinstate Monica Feb 22 '16 at 14:27
• Ok u have a point here...i'm a theoretical physicist and used diff. geo mainly in the way it is used in general relativity and then Lee's definition is at least for me the more natural one :) – tired Feb 22 '16 at 14:38