Let $\mathbb{K} \subset \mathbb{T}$ be two tori acting on a topological space $X$ (with all the properties you want). We use the notations $$X_{\mathbb{T}} := (X \times E \mathbb{T}) / \mathbb{T}, \quad X_{\mathbb{K}} := (X \times E \mathbb{K}) / \mathbb{K}$$ for the homotopy quotients.

The space $E \mathbb{T}$ can be considered as $E \mathbb{K}$ if provided with the $\mathbb{K}$-action, and therefore there exists a fibration $$X_{\mathbb{K}} \overset{\mathbb{T} / \mathbb{K}}{\longrightarrow} X_{\mathbb{T}}.$$

The Serre spectral sequence associated with this fibration has $E_2$-term $$E_2^{pq} = H^p(X_{\mathbb{T}}) \otimes H^q(\mathbb{T} / \mathbb{K}),$$ and converges to $H^{p+q}(X_{\mathbb{K}})$.

I am trying to understand what the transgression map is here, and more precisely what are the images of the generators of $H^*(\mathbb{T} / \mathbb{K})$ in $H^2(B \mathbb{T})$.

Moreover, I read somewhere that this $E_2$-term is the Koszul complex of these images (of degree $2$) in the algebra $H^p(X_{\mathbb{T}})$. What does this mean ?

Thanks a lot


1 Answer 1


The images of the generators for $H^1(\mathbb{T}/\mathbb{K})$ are in some sense Chern classes. The easiest case to make sense of is the codimension one case, i.e., when $\mathbb{T}/\mathbb{K}\cong U(1)$. In this case you can think of $X_{\mathbb{K}}$ as a principal $U(1)$-bundle over $X_\mathbb{T}$. Principal $U(1)$ bundles are classified by their first Chern class, which is an element of $H^2$ of the base, or in this case $H^2(X_{\mathbb{T}})$. The differential $d_2$ sends the generator for $E_2^{01}=H^0(X_{\mathbb{T}})\otimes H^1(U(1))$ to $c_1\otimes 1\in E_2^{20}=H^2(X_{\mathbb{T}})\otimes H^0(U(1))$.

In the general case, fix an isomorphism from $\mathbb{T}/\mathbb{K}$ to $U(1)^m$ for some $m$. The picture is similar because a principal $U(1)^m$ bundle is naturally built from $m$ principal $U(1)$-bundles, and you have $m$ first Chern classes in $H^2(X_{\mathbb{T}})$ which are the images of the natural generators for $H^1(U(1)^m)$.

I am on shakier ground here, but Koszul resolutions are something to do with homological algebra over commutative rings. View a field $k$ as a module for the polynomial ring $k[x_1,\ldots,x_m]$, with each $x_i$ acting as multiplication by zero, and grade everything so that each $x_i$ has degree two. If you want to make a free resolution for the module $k$, the most economical choice will involve taking a free module of rank $m\choose{j}$ in degree $j$. The differential in this resolution can be described by viewing the whole resolution as a bigraded algebra $\Lambda[y_1,\ldots,y_m]\otimes k[x_1,\ldots,x_m]$, where each $y_i$ has bidegree $(0,1)$ and $x_i$ has bidegree $(2,0)$, and $d(y_i)=x_i$. This bicomplex is identical to the one that you have in the case when $X$ is a single point (i.e., the case of the path-loop fibration over $U(1)^m$). I recommend Matsumura's book `Commutative Ring Theory' if you want to learn more about homological algebra over commutative rings.


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