# Linear Spaces of a Graph, Part 3

## When are *Z* and *B* disjoint?

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This material comes from some notes for a
University of Waterloo
course taught by Herb Shank. (c. 1975)

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As another application of Theorem 3, we investigate whether *Z* and *B* are disjoint. If they are, then by adding dimensions, we have immediately that *C* is the direct sum of *Z* and *B*. It turns out that whether *Z* and *B* are disjoint depends on the relation of the number *N* of spanning trees to the characteristic of the field *F*.

**Theorem 5**.
*Z*_{F}(G) and *B*_{F}(G) are disjoint if and only if *N(G)* is not a multiple of the characteristic of *F*. Moreover, if *Z* and *B* are disjoint, the projection on *C* onto *Z* is
*N*^{ -1}(A_{T1} +
A_{T2} + ...) and that of
*C* onto *B* is
*N*^{ -1}(B_{T1} +
B_{T2} + ...).

**Proof**. By Theorem 3, using *R=I*, we know that
*A*_{T1} +
A_{T2} + ...
=
A_{T1}^{*} +
A_{T2}^{*} + ...

Suppose that *N* is not zero in *F*. Note that each *A*_{T} is the identity on *Z* and each
*A*_{T}^{*} sends cocyles to zero (the rows of the matrix representing
*A*_{T}^{*} are the columns of the matrix representing
*A*_{T} and are thus cycles, which by definition have inner product zero with any cocycle).
Thus,
*N*^{ -1}(A_{T1} +
A_{T2} + ...) preserves cycles and sends cocyles to zero.
For the projection expression for *B*, use the above and Lemma 6:
*A*_{T} + B_{T}^{*} = I.

The following proof of the converse is due to S. Maurer: Suppose that *N* is zero in *F*. It could happen that
*(A*_{T1} +
A_{T2} + ...)c is non-zero for some *c*. If so, then
*(A*_{T1} +
A_{T2} + ...)c +
(B_{T1} +
B_{T2} + ...)c = Nc = 0, so that
*(A*_{T1} +
A_{T2} + ...)c =
-(B_{T1} +
B_{T2} + ...)c is a non-zero member of
*Z(G)* intersect *B(G)*. Otherwise,
*A*_{T1} +
A_{T2} + ... is the zero linear transform and, in particular, for an edge *e*
* ( (A*_{T1} +
A_{T2} + ...)e , e ) = 0.
Since this is the number of spanning trees of *G* not containing *e*, *N(G-e)=0* and, by Lemma 3 and Corollary,
*N(G*_{e} )=0. By induction of the number of edges, there is some non-zero *d*_{1} in *Z(G-e)* intersect *B(G-e)* and some non-zero *d*_{2} in *Z(G*_{e} ) intersect *B(G*_{e} ).
By Lemma 4 and Lemma 5,

*d*_{1} in *Z(G)*,
*d*_{2}+k_{2}e in *Z(G*_{e} ) and

*d*_{2} in *B(G)*,
*d*_{1}+k_{1}e in *B(G*_{e} ).

If *k*_{1}=0, then *d*_{1} is in *Z(G)* intersect *B(G)*.

If *k*_{2}=0, then *d*_{2} is in *Z(G)* intersect *B(G)*.

Otherwise,
*(k*_{1}^{-1}d_{1}+e) + k_{2}^{-1}d_{2} is in *B(G)* and
*(k*_{2}^{-1}d_{2}+e) + k_{1}^{-1}d_{1} is in *Z(G)*.

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Last modified November 3, 2000, by S.C. Locke.