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In Renteln's, Manifolds, Tensors and Forms, p. 81, The cotangent space as a jet space$^*$, we have the following definitions

Let $f:M \to \mathbb R$ be a smooth function, $p \in M$, and $\{x^i\}$ local coordinates around $p$. We say the $f$ vanishes to first order at $p$ if $\partial f/\partial x^i$ vanishes at $p$ for all $i$.

But there is no requirement on $f(p)$ to vanish. Moreover

Inductively, for $k \ge 1$ we say that $f$ vanishes to $k$th order at $p$ if, for every $i$, $\partial f/\partial x^i$ vanishes to $(k-1)$th order at $p$, where $f$ vanishes to zeroth order at $p$ if $f(p) = 0$. Put another way, $f$ vanishes to $k$th order at $p$ if the first $k$ terms in its Taylor expansion vanish at $p$.

I'm not sure whether he meant "orders" instead of "terms" in the last sentence, or whether the "first $k$" counts from 0 or from 1, but he continues

For $k > 0$ let $M_p^k$ denote the set of all smooth functions on $M$ vanishing to $(k − 1)$th order at $p$, and set $M_p^0 := \Omega^0(M)$ and $M_p := M_p^1$. Each $M_p^k$ is a vector space under the usual pointwise operations, and we have the series of inclusions $$ M_p^0 \supset M_p^1 \supset M_p^2 \dots $$

So $M_p = M_p^1$ is the set of functions vanishing to zeroth order at $p$, i.e. all those with $f(p) = 0$. $M_p^2$ is the set of those with $\partial f/\partial x^i = 0$ but not necessarily $f(p) = 0$. So the inclusion are not correct? Finally

We now define $T_p^*M$, the cotangent space to $M$ at $p$, to be the quotient space $$ T_p^*M = M_p/M_p^2. $$

Which is the set of equivalence classes, each of which consists of functions different from each other by a function vanishing to the 1st-order (with zeroth 1st partial derivative).

Question: Is it needed or not that $f \in M_p^k, k \ge 1$ should have $f(p) = 0$ ? If not then I suppose it's possible to define the cotangent space simply as $$ T_p^*M = \Omega^0(M)/M_p^2. $$

Physor
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  • The author says there that "$x$ vanishes to zeroth order at $0$, $x^2$ vanishes to first order at $0$" and so on. So, when you count "first $k$ terms in its Taylor expansion", you need to start from zero. I believe that you have spotted inconsistencies in the text. – Yuri Vyatkin Sep 18 '21 at 12:15
  • I had to add, I don't know whether this definition (after any repair needed) would be independent of the coordinates map $\phi$ used to make sense of $\partial f/\partial x^i \equiv \partial (f\circ \phi^{-1})/\partial x^i$ – Physor Sep 18 '21 at 12:22
  • @YuriVyatkin I didn't want to mention that because that would make the question longer, but it (the question) has to do with whether theses functions in $M_p^k$ should vanish at $p$ or not – Physor Sep 18 '21 at 12:25
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    Fair enough. As I said, this text is confusing to me as well (otherwise I would post an answer). Personally, I prefer a coordinate-free way to think about that, as you may see here. – Yuri Vyatkin Sep 18 '21 at 12:28
  • I think that the language use is confusing, but since you can replace $f$ with $f-f(p)$ without affecting any derivatives, it's sort of a moot point. But I agree that a function that vanishes to first order should also vanish to zeroth order. Why don't you send the author an email? He's emeritus at Cal State Univ San Bernardino. – Ted Shifrin Sep 18 '21 at 18:13
  • @TedShifrin I may try but I don't know how they write to an author, should I mention my real name ? should I expect a reply ? – Physor Sep 18 '21 at 18:22
  • I actually alerted him to this post in an email a few seconds ago (I have known him since his undergraduate days). But if you send a polite inquiry, most authors are happy to hear from you. His webpage particularly invites communications about the book. He does have an errata list (which you should download), but this isn't on it. That aside, do you agree with my comment? It is an awkward omission, but it doesn't really affect anything as long as you make my adjustment. – Ted Shifrin Sep 18 '21 at 18:27
  • update: (some corrections) I didn't get your point in the comment about replacement exactly. But I had also proposed a definition at the end that I hope it is not affected whether one requires vanishing at $p$ or not. All needed is the quotient by the space of functions whose (at least) first derivative is zero. The result is will be equivalence classes distinguished from one another by the first derivative (understood as a linear map), and so the quotient space is as dimensional as that of linear maps from $\mathbb R^m$ into $\mathbb R$ – Physor Sep 18 '21 at 19:22

1 Answers1

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Physor is correct. The condition that $f(p)=0$ was inadvertently omitted from the definition of vanishing to first order. (I have updated the list of errata to include this.)

Also, Yuri is correct that the phrase `` …the first $k$ terms in the Taylor expansion…’’ means to start counting at $k=0$ for the constant term, $k=1$ for the linear term, etc..

I agree that the discussion in that section probably ought to have been rewritten to make it clearer. Perhaps if there’s ever a second edition…

Paul Renteln
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