Timeline for Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?
Current License: CC BY-SA 3.0
7 events
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| Mar 19, 2015 at 2:17 | history | edited | owb | CC BY-SA 3.0 |
I modify a proof to get a stronger result with SL instead of GL.
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| Mar 18, 2015 at 20:56 | comment | added | owb | Yes, you are right, the question I formulated in the comment does not make sense, and the proof I suggested in the answer shows linearity of all free groups without axiom of choice. | |
| Mar 18, 2015 at 20:35 | comment | added | YCor | First in ZF you should careful on what is the meaning of "uncountable", probably you should assume the countable (dependent?) axiom of choice. Second, you definitely have a proof in ZF that the free group over an arbitrary set $X$ is linear: consider the field $K$ of fractions over indeterminates $t_{x,i}$ indexed by $X\times\{1,2,3,4\}$ and consider for $x\in X$ the $2\times 2$ matrices $A_x$ with entries $t_{x,1},t_{x,2},t_{x,3},t_{x,4}$; then they freely generate a free subgroup of $GL_2(K)$. | |
| Mar 18, 2015 at 17:56 | comment | added | owb | I wouldn't agree that using axiom of choice is not really different from using ultraproducts: ultraproduct is a special non-obvious construction, but axiom of choice is used everywhere as an obvious statement. On the other hand, for set-theoretists the following is a natural and typical problem : "Can be proven in ZF that uncountable free groups are linear?". Probably, the answer is "no". | |
| Mar 18, 2015 at 7:54 | comment | added | YCor | You use a transcendence basis, whose existence makes use of the axiom of choice, which is not really different from using ultraproducts. However, in the field, say, of real numbers, you can construct uncountable (continuum) algebraically independent families without the use of the axiom of choice and this indeed provides free subgroups of uncountable rank in $GL_2(\mathbf{R})$. | |
| Mar 17, 2015 at 22:53 | history | edited | owb | CC BY-SA 3.0 |
I added some details in the proof of Proposition in the case when $K$ is of prime characteristic.
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| Mar 13, 2015 at 2:14 | history | answered | owb | CC BY-SA 3.0 |