another way to put it: once we know that Q is true, then we know "P OR Q" must be true. but all we know for AND is that, if 'P AND Q" will end up being false, it won't because of Q. So "P AND Q = P" in that case.
Now, how for gd sake this is true: $𝑝 → (𝑞 → 𝑟)~ and ~(𝑝 ∧ 𝑞) → 𝑟$
W/o truth table!
There is a rule which says that $(p \to r) \lor (q \to r) \equiv (p \land q) \to r$. So how this is true that $p→(q→r) ~and ~(p∧q)→r$ without using truth table?
Can we solve that (p→r)∨(q→r)≡(p∧q)→r simply by taking a row from truth table and see what it would equal to? But the problem with this approach is that it's very slow as the the row we might take might yield same results
I mean try to plug same truth values for both (p→r)∨(q→r) and for (p∧q)→r and see whether they are equivvalent or not. What do you think please?
Is this the fastest approach you could think on?
Which of the following set of expressions is not equivalent to each other? a) 𝑝 and (𝑝 ∧ 𝑞) ∨ (𝑝 ∧ ¬𝑞) b) 𝑝 → (𝑞 → 𝑟) and (𝑝 ∧ 𝑞) → 𝑟 c) (𝑝 → 𝑞) → 𝑟 and (𝑝 ∧ 𝑞) → 𝑟 d) All of the above are correct
That is the question. You should solve this in ONE MINUTE :/
There's an $S_n$-action on $T^n$ by permuting the circle factors, which gives an $S_n$-action on the representation ring $\mathrm{Rep}(T^n)$. This map $\mathrm{Rep}(U(n)) \to \mathrm{Rep}(T^n)$ is an isomorphism onto $\mathrm{Rep}(T^n)^{S_n}$.
@ShaVuklia Oh cool, I'm fond of that construction.
@LeakyNun We can now start computing $\mathrm{Rep}(U(n))$. $T^n$-representations are sum of 1-dimensional things, concentrated on each circle factor. So $\mathrm{Rep}(T^n) \cong \Bbb Z[x_1, x_1^{-1}, \cdots, x_n, x_n^{-1}]$.
The $S_n$-action is by permuting, and the invariant subring can be verified to be $\Bbb Z[e_1, \cdots, e_n, e_n^{-1}]$.
This is $\mathrm{Rep}(U(n))$.
I would now like to pin down the "augmentation ideal" of $\mathrm{Rep}(U(n))$. For any $G$, there's a map $\mathrm{Rep}(G) \to \Bbb Z$ given by $\sum a_i [V_i] \to \sum a_i$, "degree" of the virtual representation.
The kernel is the augmentation ideal $I_G$. What is this object here?
I think the last statement is wrong in the sense that $H^*(BU(n)) \cong \Bbb Z[c_1, \cdots, c_n]$ is indeed a polynomial ring, but $|c_k| = 2k$, unlike $|e_k| = \binom{n}{k}$ in $\mathrm{Rep}(U(n)) \cong \Bbb Z[e_1, \cdots, e_n, e_n^{-1}]$.
I don't know how to describe the augmentation ideal in $\Bbb Z[e_1, \cdots, e_n, e_n^{-1}]$ neatly, and I am very skeptical that you can complete at a typical ideal to get a polynomial ring back.
I guess completion at eg, $(e_n)$ would indeed give you back a polynomial ring, say. But the augmentation ideal is some monster.
Gah my misunderstanding is weirder than I thought it should be.
Here are all the statements: (1) $\widehat{\mathrm{Rep}(G)}^{I_G} \cong K^*(BG)$ (2) $K^*(BG) = \bigoplus_{d \geq 0} K^d(BG)$ (3) $K^d(BG) \cong K^{d+2}(BG)$ (4) $K^0(BG) \otimes \Bbb C \cong \bigoplus_{d \, even} H^d(BG; \Bbb C)$ (5) $K^1(BG) \otimes \Bbb C \cong \bigoplus_{d \, odd} H^d(BG; \Bbb C)$
$K^*(-)$ is a graded ring, 2-periodic and when I feed in $BG$ it's known upto torsion in the $0$ and $1$ grades
If $G = U(n)$, $H^*(BU(n); \Bbb C) = \Bbb C[c_1, \cdots, c_n]$ where $|c_k| = 2k$. Fact.
So there is nothing in the odd parts and everything in the even parts.
This says $K^0(BU(n)) \otimes \Bbb C \cong \Bbb C[c_1, \cdots, c_n]$ as groups.
One second. $K^n(-)$ is individually a ring too, and the above isomorphism is a ring isomorphism
So what is $K^*(-)$ damnit
There's some extra structure
Maybe one should be careful about what the 2-periodicity map $K^d(X) \cong K^{d+2}(X)$ is
$K^{2}(X) = K^0(X \wedge S^2)$ and $X \wedge S^2$ is a basepoint version of $X \times S^2$. I can tensor a bundle over $X$ with the canonical bundle of $S^2$, and I guess that's the periodicity map $K^0(X) \to K^2(X) \to \cdots$
