Timeline for Sign of the permutation when I show that $\star{\star w}= (-1)^{n(n-k)} w$ for the Hodge operator
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3 at 21:28 | comment | added | Wrloord | Yes, this is what I wanted to clarify. Thank you very much for your valuable intervention! | |
| Jun 3 at 21:20 | comment | added | mme | Just to complete the discussion: for me, $\sigma_I$ is the name of a permutation, whose sign you denote by $(-1)^I$. By definition, we have $$\sigma_{I^c}(m) = \begin{cases} j_m & 1 m \le n-k \\ i_{m-n+k} & m > n-k \end{cases}$$ Now for $m \le k$ we have $$\sigma_{I^c}(\tau(m)) = \sigma_{I^c}(m+n-k) = i_{m+n-k-n+k} = i_m$$ and for $m > k$ we have $$\sigma_{I^c}(\tau(m)) = \sigma_{I^c}(m-k) = j_{m-k},$$ which is precisely the definition of $\sigma_I$. | |
| Jun 3 at 21:18 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing (used $\star$ instead of $\ast$ for the Hodge star operator)
|
| Jun 3 at 19:46 | comment | added | Wrloord | As far as I can see, for example for $m \leq k$, $\tau_{k,n}(m)=n-k+m$ (note $n-k+m >n-k$, then $\sigma_{I^c}=i_{(n-k+m)-(n-k)}=i_{m}$ analogous for the other case, this may be a reason, by the way thank you for your help in answering the question. | |
| Jun 3 at 18:43 | comment | added | Wrloord | From your idea I think I could deduce that $(-1)^{I} (-1)^{J}=(-1)^{k(n-k)}$, where $I, J$ are permutations resulting from applying the operator, and with this we would have the result and it seems to work correctly, but I still can't understand the following conclusion: $\sigma_{I}=\sigma_{I^{c}} \tau$ | |
| Jun 3 at 18:37 | vote | accept | Wrloord | ||
| Jun 3 at 17:29 | history | edited | LSpice | CC BY-SA 4.0 |
Spacing in $*{*dx_I}$
|
| Jun 3 at 12:04 | history | edited | mme | CC BY-SA 4.0 |
added 522 characters in body
|
| Jun 3 at 10:19 | comment | added | mme | That's the method I wanted to convey, but instead of relating the signs, I had in mind to directly relate the permutations. Because we have an equality of permutations $\sigma_I = \sigma_{I^c} \tau$, we have that $\text{sign}(\sigma_I) = \text{sign}(\sigma_{I^c}) \text{sign}(\tau)$. It then suffices to compute directly that $\tau$ is a product of $k(n-k)$ transpositions. | |
| Jun 3 at 3:35 | comment | added | Wrloord | Thanks for your help, I'll try to land the idea a bit, namely that (in my notation) $(-1)^{\sigma_I}$ and $(-1)^{\sigma_J}$ differ from being equal by the factor $(-1)^{k(n-k)}$, and hence when $(-1)^{k(n-k)}$ arises from the commutativity of the wedge product the result adjusts, that's kind of how I see it, if this is not the idea you want to convey, can you elaborate a little more on the combinatorial method? I would be extremely grateful | |
| Jun 3 at 3:31 | comment | added | mme | Direct argument: write $\tau$ for the permutation $$\tau(i) = \begin{cases} n-k+i & i \le k \\ i - k & i > k \end{cases}$$ Then $\sigma_{I} = \sigma_{I^c} \tau$. It follows that the signs of $\sigma_I$ and $\sigma_{I^c}$ differ by the sign of $\tau$, which is $(-1)^{k(n-k)}$. Basically the same argument as above. | |
| Jun 3 at 3:27 | comment | added | mme | That $\sigma(I^c) \sigma(I)$ has sign $(-1)^{k(n-k)}$ follows from this. This can also be seen directly / combinatorially. | |
| Jun 3 at 3:25 | history | answered | mme | CC BY-SA 4.0 |