Timeline for Thinking of a linear operator as a (1,1) tensor
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2022 at 3:03 | comment | added | roshoka | I see, thanks for the information | |
| Aug 16, 2022 at 18:13 | comment | added | Gold | This is a general pattern in Math. We often have objects which can be identified in the sense that there exists a bijective correspondence between them and that this bijective map respects whatever extra structure is in place. This leads to the notion of isomorphisms in algebra, homeomorphisms in topology, diffeomorphisms in differential geometry and so forth. | |
| Aug 16, 2022 at 18:12 | comment | added | Gold | Moreover, since the objects are elements of vector spaces, they can be combined linearly. So we ask that such a map ${\cal I}$ be linear, thereby preserving linear combinations. In that case ${\cal I}$ will be one linear isomorphism between vector spaces. | |
| Aug 16, 2022 at 18:10 | comment | added | Gold | Your question is about the identification of two, at first distinct, objects: $(1,1)$ tensors and linear operators. Mathematically such identification means that there should be a one-to-one correspondence between such objects, so that when you get a linear operator $A$ you know there will be one and exactly one $(1,1)$ tensor $\widetilde{A}$ corresponding to it and vice-versa. This means there should be such an invertible map ${\cal I}$ with inverse ${\cal I}^-$. | |
| Aug 16, 2022 at 4:35 | comment | added | roshoka | It seems like before the components part the important thing was to find $\mathcal{I}^{-1}$. What is the significance of that? Find the inverse in the first place? | |
| Aug 16, 2022 at 4:14 | vote | accept | roshoka | ||
| Aug 13, 2022 at 19:08 | history | answered | Gold | CC BY-SA 4.0 |