Timeline for Dimensions of dual vector spaces
Current License: CC BY-SA 3.0
21 events
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| S Feb 20, 2016 at 19:53 | history | bounty ended | CommunityBot | ||
| S Feb 20, 2016 at 19:53 | history | notice removed | CommunityBot | ||
| Feb 17, 2016 at 15:35 | comment | added | Pace Nielsen | @JeremyRickard That essentially sums up what I know as well. In every example I've tried where $E=F$ and $_EV$ is infinite dimensional, the $E$-dimension on $V^{\ast}$ is no smaller. | |
| Feb 17, 2016 at 11:19 | comment | added | Jeremy Rickard | Just to expand on a previous comment. If $E$ and $F$ are fields and $V$ is an $E$-$F$-bimodule, infinite-dimensional over $F$, then $V^\ast=\operatorname{Hom}_F(V,F)$ is an $F$-$E$-bimodule whose $F$-dimension is strictly greater than that of $V$. So if $E=F$ and $V$ answers the question, we'd need the $E$-dimension of $V^\ast$ to be strictly less than that of $V$. Even without $E=F$ I don't know if this is possible, although it is possible for the $E$-dimensions of $V$ and $V^\ast$ to be equal (and infinite). | |
| S Feb 12, 2016 at 18:05 | history | bounty started | Pace Nielsen | ||
| S Feb 12, 2016 at 18:05 | history | notice added | Pace Nielsen | Draw attention | |
| Feb 4, 2016 at 19:37 | comment | added | Manny Reyes | @PaceNielsen Thanks, that's exactly what I was missing. | |
| Feb 4, 2016 at 18:19 | comment | added | Pace Nielsen | In other words, while we can understand the dimension of $V^{\ast}$ on one side very well (the side opposite the one with which we take homomorphisms), the dimension on the other side is a complete mystery to me. | |
| Feb 4, 2016 at 18:16 | comment | added | Pace Nielsen | @MannyReyes Note that the side we are taking the $F$-homomorphisms on matters, so if you take opposites, you must also take the homomorphisms on the other side. So, let's use right homomorphisms as in the post. Then it is forced to have $\dim(V_F)<\dim(_FV)$. | |
| Feb 4, 2016 at 16:46 | comment | added | Manny Reyes | Pace, I feel that I must be misunderstanding something, but the answer seems to be negative and to follow from your fact about one-sided duals. Please let me know if I've made a silly error. The dimension on one of the sides of $V$ must be $\geq$ than the dimension on the other side; WLOG (take opposites if necessary) say $\dim(V_F) \geq \dim({}_F V)$. But then $\dim({}_F V^*) > \dim(V_F) \geq \dim({}_F V)$ so that the dimensions of $V$ and $V^*$ can't agree on both sides. | |
| Feb 4, 2016 at 15:00 | comment | added | Pace Nielsen | @JeremyRickard I do not know even that much. (And thanks for spotting the typos!) | |
| Feb 4, 2016 at 14:59 | history | edited | Pace Nielsen | CC BY-SA 3.0 |
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| Feb 4, 2016 at 14:15 | comment | added | Jeremy Rickard | Do you know examples of bimodules $_EV_F$ for fields $E$ and $F$ where the $E$-dimension of $\operatorname{Hom}_F(V,F)$ is strictly Iess than the $E$-dimension of $V$? By the way, I think that in the second paragraph $D$ should be $F$? | |
| Feb 3, 2016 at 22:30 | comment | added | Pace Nielsen | @Gro-Tsen: Regarding your first comment, the answer is yes. As long as the cardinalities of $V$ and $V^{\ast}$ are equal, and bigger than the cardinality of the prime subfield of $F$, you should get isomorphic abelian groups. Regarding your second comment, it isn't too difficult to get a bimodule $V$ where on the right the dimension is countable and on the left it is continuum. I'll post such an example above. | |
| Feb 3, 2016 at 22:29 | history | edited | Pace Nielsen | CC BY-SA 3.0 |
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| Feb 3, 2016 at 22:09 | comment | added | Gro-Tsen | Also: I suspect one can — and I suspect you know how to — manufacture an example where $V$ and $V^*$ have the same dimension on one side. If so, you should maybe state this, and give an example. (And if not, you should maybe add this as an easier or preliminary question.) | |
| Feb 3, 2016 at 22:01 | comment | added | Gro-Tsen | Is this the sort of example of the "perhaps less well known" fact? If $F$ is the field of rational functions over $\mathbb{Q}$ in $2^{\aleph_0}$ indeterminates, which has dimension $2^{\aleph_0}$ over $\mathbb{Q}$ (I think), and if $V$ has dimension $\aleph_0$ over $F$ then $V^*$ has dimension $2^{\aleph_0}$ over $F$, so they both have dimension $2^{\aleph_0}$ over $\mathbb{Q}$ (so they have the same cardinal and abelian group structure). Right? | |
| Feb 3, 2016 at 20:49 | history | edited | Pace Nielsen | CC BY-SA 3.0 |
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| Feb 3, 2016 at 18:02 | history | edited | Pace Nielsen | CC BY-SA 3.0 |
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| Jan 28, 2016 at 20:03 | history | edited | Pace Nielsen |
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| Jan 27, 2016 at 20:19 | history | asked | Pace Nielsen | CC BY-SA 3.0 |