relationship For this to make sense, we must find a differential on each and verify that it leads to homology isomorphic to . The differential is defined by restricting the original differential defined on to the subobject . It is straightforward to check that the homology of with respect to this differential is , so this gives a spectral sequence. Unfortunately, the differential is not very explicit. Determining differentials or finding ways to work around them is one of the main challenges to successfully applying a spectral sequence. Spectral sequences constructed with this method Hodge–de Rham spectral sequence Spectral sequence of a double complex Can be used to construct Mixed Hodge structures The spectral sequence of a double complex Another common spectral sequence is the spectral sequence of a double complex. A double complex is a collection of objects Ci,j for all integers i and j together with two differentials, d I and d II. d I is assumed to decrease i, and d II is assumed to decrease j. Furthermore, we assume that the differentials anticommute, so that d I d II + d II d I = 0. Our goal is to compare the iterated homologies and . We will do this by filtering our double complex in two different ways. Here are our filtrations: To get a spectral sequence, we will reduce to the previous example. We define the total complex T(C•,•) to be the complex whose n'th term is and whose differential is d I + d II. This is a complex because d I and d II are anticommuting differentials. The two filtrations on Ci,j give two filtrations on the total complex: To show that these spectral sequences give information about the iterated homologies, we will work out the E0, E1, and E2 terms of the I filtration on T(C•,•). The E0 term is clear: where . To find the E1 term, we need to determine d I + d II on E0. Notice that the differential must have degree −1 with respect to n, so we get a map Consequently, the differential on E0 is the map Cp,q → Cp,q−1 induced by d I + d II. But d I has the wrong degree to induce such a map, so d I must be zero on E0. That means the differential is exactly d II, so we get To find E2, we need to