Consider the commutative diagram in which the top row is...

Consider the commutative diagram in which the top row is exact, where $\operatorname{im}\left(A_{p+1} \rightarrow A_{p}\right)=K_{p}=\operatorname{ker}\left(A_{p} \rightarrow A_{p-1}\right)$. For each $p$ and module $C$, the exact sequence $0 \rightarrow K_{p} \rightarrow A_{p} \rightarrow K_{p-1} \rightarrow 0$ gives the long exact sequence $\operatorname{Tor}_{q}\left(C, K_{p}\right) \stackrel{\beta}{\rightarrow} \operatorname{Tor}_{q}\left(C, A_{p}\right) \stackrel{y}{\rightarrow} \operatorname{Tor}_{q}\left(C, K_{p-1}\right) \stackrel{\alpha}{\rightarrow} \operatorname{Tor}_{q-1}\left(C, K_{p}\right) .$ If $D_{p, q}=\operatorname{Tor}_{q}\left(C, K_{p}\right)$ and $E_{p, q}=\operatorname{Tor}_{q}\left(C, A_{p}\right)$, prove that $\alpha, \beta$, and $\gamma$ are bigraded maps with respective bidegrees $(1,-1),(0,0)$, and $(-1,0)$ and that ( $D, E, \alpha, \beta, \gamma$ ) is an exact couple. (Notice that this exact couple does not arise from a filtration.)

Consider the commutative diagram in which the top row is exact,
where $\operatorname{im}\left(A_{p+1} \rightarrow A_{p}\right)=K_{p}=\operatorname{ker}\left(A_{p} \rightarrow A_{p-1}\right)$. For each $p$ and module $C$, the exact sequence $0 \rightarrow K_{p} \rightarrow A_{p} \rightarrow K_{p-1} \rightarrow 0$ gives the long exact sequence
$\operatorname{Tor}_{q}\left(C, K_{p}\right) \stackrel{\beta}{\rightarrow} \operatorname{Tor}_{q}\left(C, A_{p}\right) \stackrel{y}{\rightarrow} \operatorname{Tor}_{q}\left(C, K_{p-1}\right) \stackrel{\alpha}{\rightarrow} \operatorname{Tor}_{q-1}\left(C, K_{p}\right) .$
If $D_{p, q}=\operatorname{Tor}_{q}\left(C, K_{p}\right)$ and $E_{p, q}=\operatorname{Tor}_{q}\left(C, A_{p}\right)$, prove that $\alpha, \beta$, and $\gamma$ are bigraded maps with respective bidegrees $(1,-1),(0,0)$, and $(-1,0)$ and that ( $D, E, \alpha, \beta, \gamma$ ) is an exact couple. (Notice that this exact couple does not arise from a filtration.)

Key Concepts

Bigraded Maps

Bigraded maps are maps between bigraded modules that shift the degrees in each grading by specified amounts. The bidegree of a map is an ordered pair of integers indicating how the map increases or decreases the first and second grading, respectively. When dealing with spectral sequences or complexes with two gradings, understanding these shifts is essential for tracking the movement of elements between different bidegree components under the action of the maps.

Tor Functor

The Tor functor is a central concept in homological algebra that measures how non-flat a module is. It is constructed from a projective resolution of a module and computes derived functors of the tensor product. In contexts such as exact sequences, the Tor functor allows us to pass from short exact sequences of modules to long exact sequences in homology, capturing information about the relationships between different modules through their derived tensor products.

Exact Couple

An exact couple is an algebraic structure consisting of two modules (or graded modules) and three maps between them that fit into a triangle, with each vertex of the triangle given by a module and each arrow being one of the maps. The exactness condition at each vertex ensures that the image of one map equals the kernel of the following map. Exact couples are frequently used as a tool to construct spectral sequences, where the repeated iteration of the structure’s maps gives rise to successive pages of the sequence.

Long Exact Sequence in Homological Algebra

Long exact sequences arise naturally from short exact sequences of modules when applying a left or right exact functor. In homological algebra, such sequences are vital because they relate the homology (or derived functors) of different modules. They encode the failure of a certain functor to be exact by introducing connecting homomorphisms, which in this context help to define the maps (like ? and ?) in the exact couple, while prescribing how degrees shift between the terms in the sequence.

Transcript

00:01 Hello there.

00:01 In this exercise we have we need to prove the following term but we're going to follow some steps so the terms technically say that if you have a two by two matrix you can write this matrix as as a multiplication of the following matrices here that you have the matrix p which is invertible and this matrix in the middle is composed by the coefficients that appear on the eigenvalues of the matrix a and the matrix p is composed by the columns of this matrix are composed by the real part of one of the eigenvectors and the imaginary part of the same eigenvector.

00:40 And the important thing of this theorem is that v is different from zero.

00:47 Okay? so we need to prove this.

00:54 So the first part to prove this is that we're going to use the notation m for the matrix the period in the middle that means that m it's going to be equals to a minus b b a now we need to use here the fact that a x is equals to lambda x to show that a x here here so i need more space i'm really sorry for that so a x equals to lambda x so based on this we need to to show that this implies that a x is equal to a u plus v v plus i v minus v where u is equals to the the imaginary part of x okay so let's start with this part we know that lambda in this case for this matrix in particular is equals to a minus b i and x will be equals to u plus i v okay so if we take into account this we can see that a x is equal to the multiplication here of a minus b i with the vectors u plus i v and you can observe that this will result into a u plus v v plus i plus i minus b u and plus here is plus a v okay as easy as that then we need to show that we can write okay so now we need to see what happened with the left hand side here okay so let's remember that x is equals to you plus iv so the left hand side of this expression would look like a u plus i a v is equal here to this part a u plus v v v plus a v and here we need to equate the real part and the imaginary parts and from that we observe that a u a times u is equals to a u plus v v and av is equals to minus v u plus a v now why these two expressions are important well if we multiply a would be this matrix b that let's remember that p is formed in this case by u and b because u corresponds to the real part of the eigenvector and v corresponds to the imaginary part of the eigenvector okay so ap will be the the the vector that has as columns a d u and a v and here we have the corresponding expressions so this is equals to the matrix composed by the vector the column vector a u plus v v together with the column v v minus u minus v u plus a v and now let's see that the matrix p m is the same so the matrix p is composed by u1 u 2 v1 v2 okay because this has a structure of uv okay and the matrix m is the matrix a minus b b a and you can observe that this multiplication is equal to a u1 plus v v1 a u2 plus v v2 minus v u1 plus a v1 and minus v u2 plus a v2 but this is equal to a u plus a v2 so these two a u plus v v v minus v u plus a v so these two expressions here are the same and that means that ap is equals to mp now that's the first part the last part is that if we show that p is invertible then immediately we have that a is equals to p mp inverse so here is not here is p m yeah sorry here is pn, yeah, that makes more sense.

07:37 Okay, so if p is invertible, we have the relation that we want to prove.

07:43 So we're going to show this by contradiction.

07:47 So let's assume that p is not invertible.

08:03 So if p is not invertible, that means that the columns are a linear, one of the columns is a linear combination of the other one...

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