I have a hard time understanding the definitions of the $\flat$ and $\sharp$ operators given in Lee's book on Riemannian manifolds:

The flat and sharp operators can be applied to tensors of any rank, in any index position, to convert tensors from covariant to contravariant or vice versa. Formally, this operation is defined as follows: if $F$ is any $(k,l)$-tensor and $i\in\{1,\ldots,k+l\}$ is any covariant index position for $F$ (meaning that the $i$-th argument is a vector, not a covector), we can form a new tensor $F^\sharp$ of type $(k+1,l-1)$ by setting $$F^\sharp(\alpha_1,\ldots,\alpha_{k+l})=F(\alpha_1,\ldots,\alpha_{i-1},\alpha_i^\sharp,\alpha_{i+1},\ldots,\alpha_{k+l})$$ whenever $\alpha_1,\ldots,\alpha_{k+l}$ are vectors or covectors as appropriate. In any local frame, the components of $F^\sharp$ are obtained by multiplying the components of $F$ by $g^{pq}$ and contracting one of the indices of $g^{pq}$ with the $i$-th index of $F$. Similarly, if $i$ is a contravariant index position, we can define a $(k-1,l+1)$-tensor $F^\flat$ by $$F^\flat(\alpha_1,\ldots,\alpha_{k+l})=F(\alpha_1,\ldots,\alpha_{i-1},\alpha_i^\flat,\alpha_{i+1},\ldots,\alpha_{k+l}).$$ In components, it is computed by multiplying by $g_{pq}$ and contracting.

Caution: The term "tensor" in the quote should refer to a tensor field. Similarly, as you can tell from the context, the terms "vector" and "covector" have their meanings understood as fields.

The crux of the matter: Take $\sharp$ as an example. What if $i=k+l$? In this case, $\alpha_i$ is a vector field, and I don't see any possibilities of assigning a meaning to $\alpha_i^\sharp$. Any help would be great. Thank you.


1 Answer 1


Suppose $F$ is a $(k,l)$-tensor field on $M$, and the factors are arranged so that the first $k$ entries of $F$ are covector fields and the last $l$ entries are vector fields, so once fully evaluated it looks like \begin{align} F(\underbrace{\text{covector fields}}_{\text{$k$ of them}}, \underbrace{\text{vector fields}}_{\text{$l$ of them}}) \end{align} Now, one can consider a tensor field of type $(k+1,l-1)$, denoted say as $F^{\sharp}$. This is of course an abuse of notation since the notation doesn't specify in which slot the $\sharp$ is acting. Nevertheless, trying to indicate this within the notation (eg $F^{\sharp, k+l}$ or some other nonsense like this) is way too cumbersome, so it's better to just say it in words. For us, let us suppose $F^{\sharp}$ has the argument structure as \begin{align} F^{\sharp}(\underbrace{\text{covector fields}}_{\text{$k$ of them}}, \underbrace{\text{vector fields}}_{\text{$l-1$ of them}}, \underbrace{\text{covector field}}_{\text{$1$ of them}}) \end{align} Notice that $F^{\sharp}$ here is eating a total of $k+1$ covector fields and $l-1$ vector fields and is thus a tensor field of type $(k+1,l-1)$. In terms of explicit formulas, let $\omega_1,\dots, \omega_k,\omega$ be covector fields and $X_1,\dots, X_{l-1}$ be vector fields. Then, we define \begin{align} F^{\sharp}(\omega_1,\dots, \omega_k, X_1,\dots, X_{l-1}, \omega):=F(\omega_1,\dots, \omega_k, X_1,\dots, X_{l-1}, \omega^{\sharp}) \end{align} This definition makes perfect sense because $\omega$ is a covector field, so $\omega^{\sharp}$ is a vector field, and is thus being fed into the last entry of $F$, which we have declared above to be a vector-field slot.

  • $\begingroup$ Thank you. So the thing is, the entries of $F^\sharp$ does NOT necessarily appear in the same order as $F$: covector fields may come after vector fields. Is that correct? $\endgroup$
    – Boar
    Sep 17, 2021 at 4:55
  • 2
    $\begingroup$ @Steve it is entirely up to you how you want to define things. I could have considered the tensor field $T(\omega_1,\dots, \omega_k,\omega,X_1,\dots, X_{l-1}):=F(\omega_1,\dots, \omega_k, X_1,\dots, X_{l-1},\omega^{\sharp})$. Clearly, $T$ and $F^{\sharp}$ as I have defined in my answer are different maps. But they're only different up to a permutation of their domains. The advantage of not "reordering" the domain so much (i.e using $F^{\sharp}$ as I have defined) is that we can simply speak of "raising/lowering" that specific index/slot. $\endgroup$
    – peek-a-boo
    Sep 17, 2021 at 4:57
  • $\begingroup$ Thank you. Your explanation is informative. $\endgroup$
    – Boar
    Sep 18, 2021 at 1:47

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