Timeline for Is the type $(1,1)$ Kronecker delta tensor, $\delta_a^{\,\,b}$ equal to the trace of the identity matrix or always $1$ when $a=b$ and zero otherwise?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 24, 2023 at 6:43 | vote | accept | CommunityBot | ||
| Oct 12, 2022 at 21:16 | history | edited | Didier | edited tags | |
| Oct 12, 2022 at 8:30 | comment | added | J.G. | The trace is $\delta_a{}^a$. | |
| Oct 12, 2022 at 8:27 | answer | added | Didier | timeline score: 1 | |
| Oct 12, 2022 at 7:59 | comment | added | Didier | Then (3) is an expression for $n \cdot A^{ij}$, where $n$ is the dimension, as you noticed, and is not related to $A_{ij}$ in any way. What is your specific question then at this precise point? | |
| Oct 12, 2022 at 7:57 | comment | added | user759380 | @Didier Hi there, I'm not trying to pass from $(2)$ to $(3)$, $(2)$ is not related to $(3)$ (as far as I can tell). Those expressions $(1-3)$ are simply possible answers to what $A_{ij}$ is equal to. | |
| Oct 12, 2022 at 7:50 | history | edited | user759380 | CC BY-SA 4.0 | Fixed grammar and typos in vector V |
| Oct 12, 2022 at 7:45 | comment | added | Didier | I do not get how you are passing from (2) to (3). I think you are misinterpreting something at this precise point | |
| Oct 12, 2022 at 7:45 | history | edited | user759380 | CC BY-SA 4.0 | Improved title wording |
| Oct 12, 2022 at 4:38 | history | asked | user759380 | CC BY-SA 4.0 |