Timeline for Why some operations on tensors don't give a tensor? [closed]
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| when toggle format | what | by | license | comment | |
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| Aug 29, 2020 at 1:04 | review | Reopen votes | |||
| Aug 29, 2020 at 8:27 | |||||
| Aug 29, 2020 at 0:43 | history | closed |
Michael Renardy Steven Landsburg abx Ben McKay Konstantinos Kanakoglou |
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| Aug 28, 2020 at 13:23 | comment | added | Deane Yang | For me it’s easier to understand if you work with an abstract vector space. Then there are no coordinates already defined. A tensor, such as the gradient, is well defined without any use of coordinates.. You get the formula you wrote only after you choose coordinates, say, by choosing a basis. But since the second formula requires coordinates to define and changes meaning in different coordinates, it can’t be a tensor. | |
| Aug 28, 2020 at 12:43 | comment | added | Konstantinos Kanakoglou | @user782220, i think that the comment from the other site, essentially says the same thing (although not in a clear manner): that your g is not actually a function. So $(\partial_x f , \partial_x f , \partial_x f)$ is not really a 1-form (a (0,1) tensor). Does this answer your question? | |
| Aug 28, 2020 at 4:22 | comment | added | user782220 | @DeaneYang V is meant to be R^3. | |
| Aug 28, 2020 at 4:21 | comment | added | user782220 | @KonstantinosKanakoglou That g is well defined is exactly the whole point of my post. See the comment by Stinking Bishop in math.stackexchange.com/questions/3801095/… where he elaborates on this. | |
| Aug 28, 2020 at 3:59 | comment | added | Deane Yang | Is $\mathbf V$ an abstract vector space or is it $\mathbb{R}^n$? | |
| Aug 28, 2020 at 3:55 | comment | added | Konstantinos Kanakoglou | The behaviour of tensors (1-forms in your case) under coordinate changes is a necessary condition for your g functions to be well defined (that is: independent of the coordinate system adopted to describe the vector space V=R^3). | |
| Aug 28, 2020 at 3:36 | comment | added | Konstantinos Kanakoglou | Your $g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3$ formula is not a well defined function $g:\mathbf{V} \to \mathbf{R}$. | |
| Aug 28, 2020 at 3:13 | comment | added | Michael Engelhardt | Indeed, a physicist would balk at more than one of the premises. First, that "$\partial_{x} f$ is just a scalar". It's not, it's the first component of a vector, which is not the same thing. But, let's indulge and interpret this as "in some fixed coordinate system, I have calculated $h=\partial_{x} f $, and I declare this $h$ that I have obtained a scalar." But then, secondly, $(h,h,h)$ is not a vector - after all, it doesn't change upon transformation. You don't get to "hit it with the usual transformation", you must show that it behaves according to that transformation - which it doesn't. | |
| Aug 28, 2020 at 2:58 | history | edited | user782220 | CC BY-SA 4.0 |
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| Aug 28, 2020 at 2:33 | review | Close votes | |||
| Aug 29, 2020 at 0:46 | |||||
| Aug 28, 2020 at 2:21 | comment | added | LSpice | One reason for a lack of authoritative answer could be want of a definition. Any of your $g$s, which is on $V^3$, not $V$, is (canonically identified with an element of $(V^*)^{\otimes3}$, so a mathematician might say (as you do) that it is a tensor. What $g$ you get, and hence with which element it is identified, depends on the coördinates used to take the partial derivative of $f$, so a physicist might care to argue that it is not a tensor. To answer your question of who is right, we need a definition. What definition do you want to use? | |
| Aug 28, 2020 at 2:05 | review | First posts | |||
| Aug 28, 2020 at 7:56 | |||||
| Aug 28, 2020 at 2:00 | history | asked | user782220 | CC BY-SA 4.0 |