The lemma goes like this (in author's notation, $\pi: V\to X$ is the projection map; if $f: Y\to X$, and $W$ is a vector bundle over $X$, then $f^* W$ is the pullback of $W$ over $Y$):

Let $X$ be a compact Hausdorff space, suppose that $V$ is a Hermitian vector bundle over $X$, and let $k$ be a natural number. Define $\epsilon: \bigwedge^k(\pi^* V)\to \bigwedge^{k+1}(\pi^*V)$ by setting $\epsilon(v, \omega)=(v, v\wedge \omega)$ for all $v\in V$ and $\omega\in \bigwedge^k(\pi^*V)$. For each element $v$ of $V$ that appears in a wedge product, let $\hat{v}$ indicate omission of $v$. Then $$ \epsilon^*(v, v_1\wedge...\wedge v_k)=(v, \sum_{j=1}^{k+1}(-1)^{j+1}\langle v_j, v_1\rangle v_1\wedge ...\wedge \hat{v}_j\wedge ...\wedge v_{k+1}) $$ for all simple wedges in $\bigwedge^k(V)_v$.

My questions:

  1. What is $\bigwedge^k(V)_v$?

  2. Following the author's notation, $\epsilon^*$ is supposed to be the pullback functor, and as such should map vector bundles over $\bigwedge^{k+1}(\pi^*V)$ to vector bundles over $\bigwedge^k(\pi^*V)$. But in the lemma it apparently maps from $\bigwedge^k(\pi^*V)$ as a space to $\bigwedge^k(\pi^*V)$ as a space? How do I make sense of this?

  3. $\omega\in \bigwedge^k(\pi^*V)$ is a typo - it should be $\omega\in\bigwedge^k(V)$, right?


2 Answers 2


I believe there are a number of typos in the passage you are quoting (presumably anyway, since I don't have access to the book--so some of this is guesswork):

First, $\epsilon$ is supposed to be a map $\bigwedge^k V \to \bigwedge^{k+1}V$ of vector bundles over $X$, and for a fixed $v \in V$ comes from the map $\omega \mapsto v \wedge \omega$. Then $\epsilon^*$ is actually supposed to be $\pi^*\epsilon$, the map induced by the pullback via $\pi$ (since $\pi^*$ is a functor, given a map of vector bundles over $X$ we obtain a map of vector bundles over $V$), hence is a map $\bigwedge^{k+1}\pi^*V = \pi^*(\bigwedge^{k+1}V) \to \pi^*(\bigwedge^{k}V) = \bigwedge^k \pi^*V$. It has the formula you provided, except for the fact that there is a typo and it should start with $(v, v_1 \wedge \ldots \wedge v_{k+1})$ (otherwise the right-hand side of the formula makes no sense).

Then, the last few words should read "...simple wedges in $\bigwedge^{k+1}(\pi^*V)_v$," which is the fiber of $\bigwedge^{k+1}(\pi^*V)$ over $v$.

  • $\begingroup$ You are right, $\bigwedge^k(V)_v$ is the fiber. Thank you. However, I don't think $\epsilon^*$ should be $\pi^*\epsilon$ - both vector bundles $\bigwedge^k(\pi^*V)$ and $\bigwedge^{k+1}(\pi^*V)$ are over $V$ (not $X$), and as such cannot be pulled back along the map $\pi: V\to X$. If it helps, the proof goes as follows: $\endgroup$ Feb 18, 2023 at 23:18
  • $\begingroup$ 1) Show that the following formula holds: $\langle\epsilon(v, v_1\wedge ...\wedge v_k), v_1'\wedge ...\wedge v'_{k+1}\rangle=\langle v_1\wedge ...\wedge v_{k+1}, \sum_{j=1}^{k+1}(-1)^{j+1}\langle v_j', v\rangle v_1'\wedge...\wedge \hat{v}_j'\wedge ...\wedge v'_{k+1}\rangle$ 2) From this, "the desired formula follows" - although I still can't imply the meaning of $\epsilon^*$, its definition seems to have something to do with the metric properties of $\epsilon$... $\endgroup$ Feb 18, 2023 at 23:22
  • $\begingroup$ As a clarification, I think the author has a fixed basis $\{v_1, ..., v_{n}\}$ in mind. That's where $v_{k+1}$ can come from, without $\epsilon$ taking it in as an argument. It actually seems vital to the proof that $\epsilon$ maps from $\bigwedge^k(\pi^*V)$, and not $\bigwedge^{k+1}(\pi^*V)$, as otherwise $\langle\epsilon(v, v_1\wedge...\wedge v_k\wedge v_{k+1}), v_1'\wedge ...\wedge v_{k+1}'\rangle=\langle v\wedge v_1\wedge...\wedge v_k\wedge v_{k+1}, v_1'\wedge ...\wedge v_{k+1}'\rangle=0$ $\endgroup$ Feb 18, 2023 at 23:27
  • $\begingroup$ I think my point was that I thought $\epsilon$ was supposed to be a map of bundles over $X$ since $\bigwedge^k V$ is a bundle over $X$. I know that $\bigwedge^k \pi^*V$ is a bundle over $V$, but it's obtained by taking the pullback of $\bigwedge^k V$ by $\pi$, so my assumption was that $\epsilon^*$ was $\pi^*\epsilon$ for $\epsilon$ a map of bundles over $X$ (note my defn of $\epsilon$ is for bundles over $X$). But of course the author is indeed defining an $\epsilon$ as a map of bundles over $V$--I just couldn't make sense of $\epsilon^*$ in that case. $\endgroup$
    – kamills
    Feb 18, 2023 at 23:49
  • $\begingroup$ But, anyway, fair enough. I wasn't aware of a fixed basis of $V$, and the desired consequences. It seems your text has a clarity issue, perhaps. Maybe you can find somewhere else in the book that this sort of construction is repeated, or where this lemma is used, to iron out some of the meanings? $\endgroup$
    – kamills
    Feb 18, 2023 at 23:51
  1. $\bigwedge^k(V)_v$ is the fiber of the vector bundle $\bigwedge^k(V)$ over $v$.

  2. $\epsilon^*: \bigwedge^{k+1}(\pi^*V)\to \bigwedge^k(\pi^*V)$ denotes the adjoint operator of $\epsilon: \bigwedge^k(\pi^*V) \to \bigwedge^{k+1}(\pi^*V)$. As such, there was a typo in the definition of $\epsilon^*$ - it takes as input $(v, v_1\wedge...\wedge v_k\wedge v_{k+1})$.

  3. That is indeed another typo in the formulation of the lemma. There were also 5 more typos in the proof of this lemma.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .