Table 4.2: Choice of D3 and D7 embedding. Filled directions are marked with a cross. Unfilled directions by a dot. 66 The key feature of figure 4.2 is the m = 0 , c = 1 . 86 curve. This shows that this geometry demonstrates chiral symmetry breaking: there is a quark condensate even if the quarks are massless! This makes the Constable-Myers Returning to the full equation of motion, we look for regular solutions that have the UV boundary condition σ = m + cρ − $^{2}$. (For the moment θ is turned off: it corresponds to the pion fields, which we will demonstrate later). A few of the solutions are plotted in figure 4.2. which has the simple solution Φ = m + cρ − $^{2}$. The two integration constants, m and c , correspond to a quark mass and a VEV for the quark bilinear ¯ qq respectively. Secondly, in the UV limit ( r → ∞ ), the equation of motion simplifies to Firstly there is a an explicit Φ → e $^{iα}$Φ symmetry. In the massless case this is the U (1)$_{A}$ symmetry on the quarks. There are two important points to be noted in equation (4.22). Applying the Euler-Lagrange equation of motion for Φ we find x$_{0}$ x$_{1}$ x$_{2}$ x$_{3}$ r ω$_{1}$ ω$_{2}$ ω$_{3}$ r$_{5}$ r$_{6}$ D3 × × × × . . . . . . D7 × × × × . . d dρ ( e $^{φ}$G ( ρ, Φ) √ 1 + | ∂$_{ρ}$ Φ | $_{2}$∂$_{ρ}$ Φ ) − 2 √ 1 + | ∂$_{ρ}$ Φ | 2 d d ¯ Φ ( e $^{φ}$G ( ρ, Φ) ) = 0 (4.22) with G ( ρ, Φ) = ρ $_{3}$(( ρ 2 + | Φ | $^{2}$)2 + 1)(( ρ 2 + | Φ | $^{2}$)2 − 1) ( ρ 2 + | Φ | $^{2}$)4 (4.23) d dρ ( ρ $_{3}$d Φ dρ ) = 0 (4.24)