Eric Auld

Oh, I'm starting to think I can just do this the way I want. I was confused. The matrix is singular iff $\langle v, u \rangle \neq -1$, and that jives with $v + \|v\|^2 u = 0$.

Hi @BalarkaSen ! Thanks for all your good answers.

In the former case $v$ is a good first singular vector. In the latter it is not, but why can't I choose a bunch of singular vectors corresponding to $\sigma = 1$ in $v^\perp$?

I know in general I can't choose my right singular vectors, but here what doesn't work? If $\mathbf{A} + \mathbf{I} + \mathbf{u}\mathbf{v}^*$, then either $\|Av\|\geq 1$ or $\|Av\|<1$.

My temptation is to take my right singular vectors to be $\mathbf{v}$ along with a basis for $\mathbf{v}^\perp$.

Suppose I take the singular value decomposition of a matrix like $\mathbf{I} + \mathbf{u}\mathbf{v}^*$.

That's awesome.

Hard to call that thing above the identity.

No, I think not. This one above I would say is not on the nose. I've sort of broken it up and put it back together.

You've just labelled everything specifically by the thing it's a degeneracy of.

Now if I do $\Gamma \circ N$, suppose I started with a $\textsf{sAb}$ called $A$. My simplicial abelian group $\Gamma \circ N (A)$ has a second level, say, that looks like $N(A_2) \oplus \texttt{all the degeneracies}$. So since as I said before, $N(A_2) \oplus \texttt{degenerate stuff} \cong A_2$, it's not too bad to convince yourself that the levels are all isomorphic, and the boundary maps if you follow them, are just the ones you started with.

No, it's not that bad! Coming in a sec

This means that $N \circ \Gamma$ of my chain complex is just literally the chain complex I started with

I ask the question, on level two say, "what are the things in $C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$ so that $d_1$ acts as zero and $d_2$ acts as zero? I claim the answer is exactly the nondegenerate term $C_2$.

If I go $N \circ \Gamma$, I'm taking that weird simplicial thing I just built from a chain complex, and then applying $N$ to it.

OK, so back to the equivalence of functors.

I can tell you a way to remember them yourself really easily if you ever want. Probably easier as a picture. Basically just think of a row of dots. $d_2 s_1 = \text{Id}$ says that doubling the first dot, relabeling the dots, and then removing the second one just gets you where you started.

The tricky thing can be to remember the simplicial identities. I guess everyone has a way of remembering them.

(remember, these are all just copies of $C_0$, just labelled differently)

$d_0 s_0$ is identity, so we're left with $s_0C_0$, and this should be the map that takes $s_0s_0C_0$ to $s_0C_0$ as the identity.

Now what about $d_0$ acting on $s_0s_0C_0$?

By simplicial identities, $d_2s_0 = s_0d_2$. So I get $s_0d_2C_1$, but all higher things act as zero on nondegenerate stuff, so this map is zero

As an example, let me look at what $d_2$ does to $s_0C_1$.

How the boundary maps act on the degenerate terms is a little more subtle.

Neither. What I said was that everything but $d_0$ acts as zero on the nondegenerate terms.

So, say I look at what $d_2$ does to the $s_0C_1$ term in level two.

Right

Yeah, that's right

what a particular boundary map does to that term. There are more than one

When I apply boundary maps to $C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$, if I focus on $s_0C_1$, say, and look at what the boundary map does to that term, it will land entirely inside of one term in the target. (Or it will be zero)

Let me rephrase that

So if I have a boundary map $d_i$, one aspect of the way we're defining them is that they'll land entirely in one summand of the target. Not spread between summands.

No, no, this is fun

Cool, I wish I had more mental energy to parse that right now.

And it will go to the $C_0$ below.

because $d_1s_0 = \text{Id}$

Yes

Now I figure out what the boundary maps do to the other terms by applying the simplicial identities.

Just like before I only took the stuff on which everything but $d_0$ acted as zero.

I declare that everything except $d_0$ shall act as zero on nondegenerate terms.

Cool

I need to define $d_0, d_1$, and $d_2$.

$C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$

Let me just retype that

Let's illustrate the boundary maps on level two, which is C2⊕s0C1⊕s1C1⊕s0s0C0C2⊕s0C1⊕s1C1⊕s0s0C0.

To that paper you just sent

Let me come back to that if I have time

The degeneracy maps for this thing are almost built in. For instance the degeneracy map $s_1$ acts level 1 by taking $C_1$ to $s_1C_1$ and $s_0C_0$ to $s_0s_0C_0$. Because $s_1s_0 = s_0s_0$.

I wouldn't be surprised. I need to understand homotopy colimits better.

Also I meant the "second" level above, not the third