Timeline for Questions about multiplicative homomorphism of $\mathbb{R}$
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2013 at 2:20 | comment | added | Todd Trimble | @woodbass: look for the check mark beside the answer, and click it. That's how you accept an answer. | |
| Jan 4, 2013 at 2:20 | vote | accept | woodbass | ||
| Jan 4, 2013 at 2:20 | vote | accept | woodbass | ||
| Jan 4, 2013 at 2:20 | |||||
| Jan 4, 2013 at 2:20 | vote | accept | woodbass | ||
| Jan 4, 2013 at 2:20 | |||||
| Jan 4, 2013 at 2:07 | history | edited | woodbass | CC BY-SA 3.0 |
edited title; edited tags
|
| Jan 4, 2013 at 0:03 | comment | added | Jim Humphreys | @woodbass: The question seems well-intended but not carefully enough constructed (including the use of "multiplication" in the header when "multiplicative" is apparently intended). Selecting the right tags is also important. Anyway, you should be able to click on a check mark to accept an answer. | |
| Jan 2, 2013 at 18:46 | comment | added | woodbass | Yes, nice answer. | |
| Jan 2, 2013 at 18:26 | comment | added | Emil Jeřábek | Regarding your last edit, $hg$ is of the same form as $h$. These endomorphisms form a group. | |
| Jan 2, 2013 at 18:21 | comment | added | Emil Jeřábek | The endomorphisms I mentioned are continuous. The other half of the story is that is every other endomorphism is neither Baire measurable nor Lebesgue measurable, and you need the axiom of choice to prove their existence, hence there is pretty much no way to describe them explicitly. Toink’s construction (with the proviso that elements of the basis can be mapped to arbitrary elements of $K$, and you have additionally a choice to map $−1$ to either $1$ or $−1$) describes all endomorphisms. | |
| Jan 2, 2013 at 18:06 | history | edited | woodbass | CC BY-SA 3.0 |
added 29 characters in body
|
| Jan 2, 2013 at 18:00 | history | edited | woodbass | CC BY-SA 3.0 |
added 123 characters in body; edited body; edited body
|
| Jan 2, 2013 at 17:54 | comment | added | woodbass | any other form of example except Tonik's? | |
| Jan 2, 2013 at 17:52 | comment | added | woodbass | Yeah. I should not have omitted these possibility. | |
| Jan 2, 2013 at 17:34 | comment | added | Emil Jeřábek | To add to Toink’s answer, there are also simple explicit examples for Q3: consider $f(x)=|x|^r$ or $f(x)=\mathrm{sgn}(x)|x|^r$ for any $r\in\mathbb R$. | |
| Jan 2, 2013 at 17:22 | comment | added | Martin Brandenburg | What do you mean by "multiplication homomorphism", perhaps just a homomorphism? And why the tag "algebraic groups"? | |
| Jan 2, 2013 at 17:12 | answer | added | Toink | timeline score: 5 | |
| Jan 2, 2013 at 17:10 | history | edited | woodbass |
edited tags
|
|
| Jan 2, 2013 at 16:52 | history | edited | woodbass | CC BY-SA 3.0 |
added 17 characters in body; edited body; deleted 4 characters in body
|
| Jan 2, 2013 at 16:44 | history | asked | woodbass | CC BY-SA 3.0 |